## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Miscellaneous Exercise 4

(I) Select the correct answers from the given alternatives.

Question 1.
The total number of terms in the expression of (x + y)100 + (x – y)100 after simplification is:
(A) 50
(B) 51
(C) 100
(D) 202
(B) 51
Hint:

Question 2.
The middle term in the expansion of (1 + x)2n will be:
(A) (n – 1)th
(B) nth
(C) (n + 1)th
(D) (n + 2)th
(C) (n + 1)th
Hint:
(1 + x)2n has (2n + 1) terms.
∴ (n + 1 )th term is the middle term.

Question 3.
In the expansion of (x2 – 2x)10, the coefficient of x16 is
(A) -1680
(B) 1680
(C) 3360
(D) 6720
(C) 3360
Hint:
(x2 – 2x)10 = x10 (x – 2)10
To get the coefficient of x16 in (x2 – 2x)10,
we need to check coefficient of x6 in (x – 2)10
∴ Required coefficient = 10C6 (-2)4
= 210 × 16
= 3360

Question 4.
The term not containing x in expansion of $$(1-x)^{2}\left(x+\frac{1}{x}\right)^{10}$$ is
(A) 11C5
(B) 10C5
(C) 10C4
(D) 10C7
(A) 11C5
Hint:

Question 5.
The number of terms in expansion of (4y + x)8 – (4y – x)8 is
(A) 4
(B) 5
(C) 8
(D) 9
(A) 4
Hint:

Question 6.
The value of 14C1 + 14C3 + 14C5 + …. + 14C11 is
(A) 214 – 1
(B) 214 – 14
(C) 212
(D) 213 – 14
(D) 213 – 14
Hint:

Question 7.
The value of 11C2 + 11C4 + 11C6 + 11C8 is equal to
(A) 210 – 1
(B) 210 – 11
(C) 210 + 12
(D) 210 – 12
(D) 210 – 12
Hint:

Question 8.
In the expansion of (3x + 2)4, the coefficient of the middle term is
(A) 36
(B) 54
(C) 81
(D) 216
(D) 216
Hint:
(3x + 2)4 has 5 terms.
∴ (3x + 2)4 has 3rd term as the middle term.
The coefficient of the middle term

= 6 × 9 × 4
= 216

Question 9.
The coefficient of the 8th term in the expansion of (1 + x)10 is:
(A) 7
(B) 120
(C) 10C8
(D) 210
(B) 120
Hint:
r = 7
t8 = 10C7 x7 = 10C3 x7
∴ Coefficient of 8th term = 10C3 = 120

Question 10.
If the coefficients of x2 and x3 in the expansion of (3 + ax)9 are the same, then the value of a is
(A) $$-\frac{7}{9}$$
(B) $$-\frac{9}{7}$$
(C) $$\frac{7}{9}$$
(D) $$\frac{9}{7}$$
(D) $$\frac{9}{7}$$
Hint:

Question 1.
Prove by the method of induction, for all n ∈ N.
(i) 8 + 17 + 26 + ….. + (9n – 1) = $$\frac{n}{2}$$ (9n + 7)
Solution:
Let P(n) ≡ 8 + 17 + 26 +…..+(9n – 1) = $$\frac{n}{2}$$ (9n + 7), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 8
R.H.S. = $$\frac{1}{2}$$ [9(1) + 7] = 8
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 8 + 17 + 26 +…..+ (9k – 1) = $$\frac{k}{2}$$ (9k + 7) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., 8 + 17 + 26 + …… + [9(k + 1) – 1]

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 8 + 17 + 26 +…..+ (9n – 1) = $$\frac{n}{2}$$ (9n + 7) for all n ∈ N.

(ii) 12 + 42 + 72 + …… + (3n – 2)2 = $$\frac{n}{2}$$ (6n2 – 3n – 1)
Solution:
Let P(n) = 12 + 42 + 72 + ….. + (3n – 2)2 = $$\frac{n}{2}$$ (6n2 – 3n – 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S.= 12 = 1
R.H.S.= $$\frac{1}{2}$$ [6(1)2 – 3(1) – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 12 + 42 + 72 +…..+ (3k – 2)2 = $$\frac{k}{2}$$ (6k2 – 3k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 12 + 42 + 72 + … + (3n – 2)2 = $$\frac{n}{2}$$ (6n2 – 3n – 1) for all n ∈ N.

(iii) 2 + 3.2 + 4.22 + …… + (n + 1) 2n-1 = n . 2n
Solution:
Let P(n) ≡ 2 + 3.2 + 4.22 +…..+ (n + 1) 2n-1 = n.2n, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(21) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 3.2 + 4.22 + ….. + (k + 1) 2k-1 = k.2k …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 3.2 + 4.22 +….+ (k + 2) 2k = (k + 1) 2k+1

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 3.2 + 4.22 +……+ (n + 1) 2n-1 = n.2n for all n ∈ N.

(iv) $$\frac{1}{3.4 .5}+\frac{2}{4.5 .6}+\frac{3}{5.6 .7}+\ldots+\frac{n}{(n+2)(n+3)(n+4)}$$ = $$\frac{n(n+1)}{6(n+3)(n+4)}$$
Solution:

Question 2.
Given that tn+1 = 5tn – 8, t1 = 3, prove by method of induction that tn = 5n-1 + 2.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation tn+1 = 5tn – 8, t1 = 3
and R.H.S. a general statement tn = 5n-1 + 2.
Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 51-1 + 2 = 1 + 2 = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S = t2 = 5t1 – 8 = 5(3) – 8 = 7
R.H.S. = t2 = 52-1 + 2 = 5 + 2 = 7
∴ L.H.S. = R.H.S.
∴ P(n) is tme for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ tk+1 = 5tk – 8 and tk = 5k-1 + 2

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
tk+1 = 5k+1-1 + 2 = 5k + 2
tk+1 = 5tk – 8 and tk = 5k-1 + 2 ……[From Step II]
∴ tk+1 = 5(5k-1 + 2) – 8 = 5k + 2
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ tn = 5n-1 + 2, for all n ∈ N.

Question 3.
Prove by method of induction
$$\left(\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right)^{n}=\left(\begin{array}{cc} 2 n+1 & -4 n \\ n & -2 n+1 \end{array}\right)$$, ∀ n ∈ N.
Solution:

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ $$\left(\begin{array}{cc} 3 & -4 \\ 1 & -1 \end{array}\right)^{n}=\left(\begin{array}{cc} 2 n+1 & -4 n \\ n & -2 n+1 \end{array}\right)$$, ∀ n ∈ N.

Question 4.
Expand (3x2 + 2y)5
Solution:
Here, a = 3x2, b = 2y, n = 5.
Using binomial theorem,

Question 5.
Expand $$\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{4}$$
Solution:

Question 6.
Find third term in the expansion of $$\left(9 x^{2}-\frac{y^{3}}{6}\right)^{4}$$
Solution:

Question 7.
Find tenth term in the expansion of $$\left(2 x^{2}+\frac{1}{x}\right)^{12}$$
Solution:

Question 8.
Find the middle term(s) in the expansion of
(i) $$\left(\frac{2 a}{3}-\frac{3}{2 a}\right)^{6}$$
Solution:
Here, a = $$\frac{2 a}{3}$$, b = $$\frac{-3}{2 a}$$, n = 6.
Now, n is even.
∴ $$\frac{\mathrm{n}+2}{2}=\frac{6+2}{2}=4$$
∴ Middle term is t4, for which r = 3.

∴ The Middle term is -20.

(ii) $$\left(x-\frac{1}{2 y}\right)^{10}$$
Solution:
Here, a = x, b = $$-\frac{1}{2 y}$$, n = 10.
Now, n is even.
∴ $$\frac{\mathrm{n}+2}{2}=\frac{10+2}{2}=6$$
∴ Middle term is t6, for which r = 5

(iii) (x2 + 2y2)7
Solution:
Here, a = x2, b = 2y2, n = 7.
Now, n is odd.
∴ $$\frac{\mathrm{n}+1}{2}=\frac{7+1}{2}=4, \frac{\mathrm{n}+3}{2}=\frac{7+3}{2}=5$$
∴ Middle terms are t4 and t5, for which r = 3 and r = 4 respectively.

∴ Middle terms are 280x8y6 and 560x6y8.

(iv) $$\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}$$
Solution:

Question 9.
Find the coefficients of
(i) x6 in the expantion of $$\left(3 x^{2}-\frac{1}{3 x}\right)^{9}$$
Solution:

(ii) x60 in the expansion of $$\left(\frac{1}{x^{2}}+x^{4}\right)^{18}$$
Solution:

Question 10.
Find the constant term in the expansion of
(i) $$\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}$$
Solution:

(ii) $$\left(2 x^{2}-\frac{1}{x}\right)^{12}$$
Solution:

Question 11.
Prove by method of induction
(i) loga xn = n loga x, x > 0, n ∈ N
Solution:

(ii) 152n-1 + 1 is divisible by 16, for all n ∈ N.
Solution:
152n-1 + 1 is divisible by 16, if and only if (152n-1 + 1) is is a multiple of 16.
Let P(n) ≡ 152n-1 + 1 = 16m, where m ∈ N.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 152n-1 + 1 is divisible by 16, for all n ∈ N.

(iii) 52n – 22n is divisible by 3, for all n ∈ N.
Solution:

Question 12.
If the coefficient of x16 in the expansion of (x2 + ax)10 is 3360, find a.
Solution:

Question 13.
If the middle term in the expansion of $$\left(x+\frac{b}{x}\right)^{6}$$ is 160, find b.
Solution:

∴ 160 = $$\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times b^{3}$$
∴ 160 = 20b3
∴ 8 = b3
∴ b = 2

Question 14.
If the coefficients of x2 and x3 in theexpansion of (3 + kx)9 are equal, find k.
Solution:

Question 15.
If the constant term in the expansion of $$\left(x^{3}+\frac{\mathrm{k}}{x^{8}}\right)^{11}$$ is 1320, find k.
Solution:

Question 16.
Show that there is no term containing x6 in the expansion of $$\left(x^{2}-\frac{3}{x}\right)^{11}$$.
Solution:

Question 17.
Show that there is no constant term in the expansion of $$\left(2 x-\frac{x^{2}}{4}\right)^{9}$$
Solution:

Question 18.
State, first four terms in the expansion of $$\left(1-\frac{2 x}{3}\right)^{-1 / 2}$$
Solution:

Question 19.
State, first four terms in the expansion of $$(1-x)^{-1 / 4}$$.
Solution:

Question 20.
State, first three terms in the expansion of $$(5+4 x)^{-1 / 2}$$
Solution:

Question 21.
Using the binomial theorem, find the value of $$\sqrt[3]{995}$$ upto four places of decimals.
Solution:

Question 22.
Find approximate value of $$\frac{1}{4.08}$$ upto four places of decimals.
Solution:

Question 23.
Find the term independent of x in the expansion of (1 – x2) $$\left(x+\frac{2}{x}\right)^{6}$$.
Solution:

Question 24.
(a + bx) (1 – x)6 = 3 – 20x + cx2 + …, then find a, b, c.
Solution:

Question 25.
The 3rd term of (1 + x)n is 36x2. Find 5th term.
Solution:

Question 26.
Suppose (1 + kx)n = 1 – 12x + 60x2 – …… find k and n.
Solution:

## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Question 1.
Show that C0 + C1 + C2 + ….. + C8 = 256
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
Putting n = 8, we get
C0 + C1 + C2 + ….. + C8 = 28
∴ C0 + C1 + C2 + ….. + C8 = 256

Question 2.
Show that C0 + C1 + C2 + …… + C9 = 512
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
Putting n = 9, we get
C0 + C1 + C2 + ….. + C9 = 29
∴ C0 + C1 + C2 + …… + C9 = 512

Question 3.
Show that C1 + C2 + C3 + ….. + C7 = 127
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
Putting n = 7, we get
C0 + C1 + C2 + ….. + C7 = 27
∴ C0 + C1 + C2 +….. + C7 = 128
But, C0 = 1
∴ 1 + C1 + C2 + ….. + C7 = 128
∴ C1 + C2 + ….. + C7 = 128 – 1 = 127

Question 4.
Show that C1 + C2 + C3 + ….. + C6 = 63
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
Putting n = 6, we get
C0 + C1 + C2 + ….. + C6 = 26
∴ C0 + C1 + C2 + …… + C6 = 64
But, C0 = 1
∴ 1 + C1 + C2 + ….. + C6 = 64
∴ C1 + C2 + ….. + C6 = 64 – 1 = 63

Question 5.
Show that C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128
Solution:
Since C0 + C1 + C2 + C3 + …… + Cn = 2n
Putting n = 8, we get
C0 + C1 + C2 + C3 + …… + C8 = 28
But, sum of even coefficients = sum of odd coefficients
∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7
Let C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = k
Now, C0 + C1 + C2 + C3 + C4 + C5 + C6 + C7 + C8 = 256
∴ (C0 + C2 + C4 + C6 + C8) + (C1 + C3 + C5 + C7) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128

Question 6.
Show that C1 + C2 + C3 + ….. + Cn = 2n – 1
Solution:
Since C0 + C1 + C2 + C3 + ….. + Cn = 2n
But, C0 = 1
∴ 1 + C1 + C2 + C3 + …… + Cn = 2n
∴ C1 + C2 + C3 + ….. + Cn = 2n – 1

Question 7.
Show that C0 + 2C1 + 3C2 + 4C3 + ….. + (n + 1)Cn = (n + 2) 2n-1
Solution:

## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)-4
Solution:

(ii) (1 – x)1/3
Solution:

(iii) (1 – x2)-3
Solution:

(iv) (1 + x)-1/5
Solution:

(v) (1 + x2)-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)-3
Solution:
(a – b)-3 = $$\left[a\left(1-\frac{b}{a}\right)\right]^{-3}$$

(ii) (a + b)-4
Solution:

(iii) (a + b)1/4
Solution:

(iv) (a – b)-1/4
Solution:
(a – b)-1/4 = $$\left[a\left(1-\frac{b}{a}\right)\right]^{\frac{-1}{4}}$$

(v) (a + b)-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)-4
Solution:

(ii) (1 + 3x)-1/2
Solution:

(iii) (2 – 3x)1/3
Solution:

(iv) (5 + 4x)-1/2
Solution:

(v) (5 – 3x)-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) $$\sqrt[3]{126}$$
Solution:

(iii) $$\sqrt[4]{16.08}$$
Solution:

(iv) (1.02)-5
Solution:

(v) (0.98)-3
Solution:

## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Question 1.
In the following expansions, find the indicated term.
(i) $$\left(2 x^{2}+\frac{3}{2 x}\right)^{8}$$, 3rd term
Solution:

(ii) $$\left(x^{2}-\frac{4}{x^{3}}\right)^{11}$$, 5th term
Solution:

(iii) $$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}$$, 7th term
Solution:

(iv) In $$\left(\frac{1}{3}+a^{2}\right)^{12}$$, 9th term
Solution:

(v) In $$\left(3 a+\frac{4}{a}\right)^{13}$$, 10th term
Solution:

Question 2.
In the following expansions, find the indicated coefficients.
(i) x3 in $$\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}$$
Solution:

(ii) x8 in $$\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}$$
Solution:

(iii) x9 in $$\left(\frac{1}{x}+x^{2}\right)^{18}$$
Solution:

(iv) x-3 in $$\left(x-\frac{1}{2 x}\right)^{5}$$
Solution:

(v) x-20 in $$\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}$$
Solution:

Question 3.
Find the constant term (term independent of x) in the expansion of
(i) $$\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$$
Solution:

(ii) $$\left(x-\frac{2}{x^{2}}\right)^{15}$$
Solution:

(iii) $$\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}$$
Solution:

(iv) $$\left(x^{2}-\frac{1}{x}\right)^{9}$$
Solution:

(v) $$\left(2 x^{2}-\frac{5}{x}\right)^{9}$$
Solution:

Question 4.
Find the middle terms in the expansion of
(i) $$\left(\frac{x}{y}+\frac{y}{x}\right)^{12}$$
Solution:

(ii) $$\left(x^{2}+\frac{1}{x}\right)^{7}$$
Solution:

(iii) $$\left(x^{2}-\frac{2}{x}\right)^{8}$$
Solution:

(iv) $$\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$$
Solution:

(v) $$\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$$
Solution:

Question 5.
In the expansion of (k + x)8, the coefficient of x5 is 10 times the coefficient of x6. Find the value of k.
Solution:

Question 6.
Find the term containing x6 in the expansion of (2 – x) (3x + 1)9.
Solution:

Question 7.
The coefficient of x2 in the expansion of (1 + 2x)m is 112. Find m.
Solution:
The coefficient of x2 in (1 + 2x)m = mC2 (22)
Given that the coefficient of x2 = 112
mC2 (4) = 112
mC2 = 28
∴ $$\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28$$
∴ $$\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28$$
∴ m(m – 1) = 56
∴ m2 – m – 56 = 0
∴ (m – 8) (m + 7) = 0
As m cannot be negative.
∴ m = 8

## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2

Question 1.
Expand:
(i) (√3 + √2)4
Solution:
Here, a = √3, b = √2 and n = 4.
Using binomial theorem,

∴ (√3 + √2)4 = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)
= 9 + 12√6 + 36 + 8√6 + 4
= 49 + 20√6

(ii) (√5 – √2)5
Solution:
Here, a = √5, b = √2 and n = 5.
Using binomial theorem,

Question 2.
Expand:
(i) (2x2 + 3)4
Solution:
Here, a = 2x2, b = 3 and n = 4.
Using binomial theorem,

(ii) $$\left(2 x-\frac{1}{x}\right)^{6}$$
Solution:
Here, a = 2x, b = $$\frac{1}{x}$$ and n = 6.
Using binomial theorem,

Question 3.
Find the value of
(i) (√3 + 1)4 – (√3 – 1)4
Solution:

(ii) (2 + √5)5 + (2 – √5)5
Solution:

Adding (i) and (ii), we get
∴ (2 + √5 )5 + (2 – √5)5 = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )
= 64 + 800 + 500
= 1364

Question 4.
Prove that:
(i) (√3 + √2)6 + (√3 – √2)6 = 970
Solution:

(ii) (√5 + 1)5 – (√5 – 1)5 = 352
Solution:

Question 5.
Using binomial theorem, find the value of
(i) (102)4
Solution:

(ii) (1.1)5
Solution:

Question 6.
Using binomial theorem, find the value of
(i) (9.9)3
Solution:

(ii) (0.9)4
Solution:

Question 7.
Without expanding, find the value of
(i) (x + 1)4 – 4(x + 1)3 (x – 1) + 6(x + 1)2 (x – 1)2 – 4(x + 1) (x – 1)3 + (x – 1)4
Solution:

(ii) (2x – 1)4 + 4(2x – 1)3 (3 – 2x) + 6(2x – 1)2 (3 – 2x)2 + 4(2x – 1)1 (3 – 2x)3 + (3 – 2x)4
Solution:

Question 8.
Find the value of (1.02)6, correct upto four places of decimals.
Solution:

Question 9.
Find the value of (1.01)5, correct upto three places of decimals.
Solution:

Question 10.
Find the value of (0.9)6, correct upto four places of decimals.
Solution:

## Maharashtra Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1

Prove by the method of induction, for all n ∈ N.

Question 1.
2 + 4 + 6 + …… + 2n = n(n + 1)
Solution:
Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 4 + 6 + ….. + 2k = k(k + 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 4 + 6 + …… + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …+ 2(k + 1)
= 2 + 4 + 6+ ….. + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …..[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n(n + 1) for all n ∈ N.

Question 2.
3 + 7 + 11 + ……… to n terms = n(2n + 1)
Solution:
Let P(n) = 3 + 7 + 11 + ……… to n terms = n(2n +1), for all n ∈ N.
But 3, 7, 11, …. are in A.P.
∴ a = 3 and d = 4
Let tn be the nth term.
∴ tn = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1
∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 1[2(1)+ 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Sept III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)
L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]
= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]
= k(2k + 1) + (4k + 4 – 1) …..[From (i)]
= 2k2 + k + 4k + 3
= 2k2 + 2k + 3k + 3
= 2k(k + 1) + 3(k + 1)
= (k + 1) (2k + 3)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

Question 3.
12 + 22 + 32 +…..+ n2 = $$\frac{n(n+1)(2 n+1)}{6}$$
Solution:
Let P(n) = 12 + 22 + 32 +…..+ n2 = $$\frac{n(n+1)(2 n+1)}{6}$$ for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 12 = 1
RHS = $$\frac{1(1+1)[2(1)+1]}{6}=\frac{6}{6}$$ = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 12 + 22 + 32 +…+ k2 = $$\frac{k(k+1)(2 k+1)}{6}$$ …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 12 + 22 + 32 + …+ n2 = $$\frac{n(n+1)(2 n+1)}{6}$$ for all n ∈ N.

Question 4.
12 + 32 + 52 + ….. + (2n – 1)2 = $$\frac{n}{3}$$ (2n – 1)(2n + 1)
Solution:
Let P(n) = 12 + 32 + 52+…..+ (2n – 1)2 = $$\frac{n}{3}$$ (2n – 1)(2n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 12 = 1
R.H.S. = $$\frac{1}{3}$$ [2(1) – 1][2(1) + 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 12 + 32 + 52 +….+(2k – 1)2 = $$\frac{k}{3}$$ (2k – 1)(2k + 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 12 + 32 + 52 + …+ (2n – 1)2 = $$\frac{n}{3}$$ (2n – 1)(2n + 1) for all n ∈ N.

Question 5.
13 + 33 + 53 + ….. to n terms = n2 (2n2 – 1)
Solution:
Let P(n) = 13 + 33 + 53 + …. to n terms = n2 (2n2 – 1), for all n ∈ N.
But 1, 3, 5, are in A.P.
∴ a = 1, d = 2
Let tn be the nth term.
tn = a + (n – 1) d = 1 + (n – 1) 2 = 2n – 1
∴ P(n) = 13 + 33 + 53 +…..+ (2n – 1)3 = n2 (2n2 – 1)

Step I:
Put n = 1
L.H.S. = 13 = 1
R.H.S. = 12 [2(1)2 – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 13 + 33 + 53 +…+ (2k – 1)3 = k2 (2k2 – 1) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 13 + 33 + 53 + … to n terms = n2 (2n2 – 1) for all n ∈ N.

Question 6.
1.2 + 2.3 + 3.4 +… + n(n + 1) = $$\frac{n}{3}$$ (n + 1)(n + 2)
Solution:
Let P(n) = 1.2 + 2.3 + 3.4 +….+n(n + 1) = $$\frac{n(n+1)(n+2)}{3}$$, for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 1.2 = 2
R.H.S. = $$\frac{1}{3}$$ (1 + 1)(1 + 2) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.2 + 2.3 + 3.4 + ….. + k(k + 1) = $$\frac{k}{3}$$ (k + 1)(k + 2) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.2 + 2.3 + 3.4 + … + n(n + 1) = $$\frac{n}{3}$$ (n + 1)(n + 2), for all n ∈ N.

Question 7.
1.3 + 3.5 + 5.7 +… to n terms = $$\frac{n}{3}$$ (4n2 + 6n – 1)
Solution:
Let P(n) = 1.3 + 3.5 + 5.7 +… to n terms = $$\frac{n}{3}$$ (4n2 + 6n -1), for all n ∈ N.
But first factor in each term, i.e., 1, 3, 5,… are in A.P. with a = 1 and d = 2.
∴ nth term = a + (n – 1)d = 1 + (n – 1) 2 = (2n – 1)
Also, second factor in each term,
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1) d = 3 + (n – 1) 2 = (2n + 1)
∴ nth term, tn = (2n – 1) (2n + 1)
∴ P(n) ≡ 1.3 + 3.5 + 5.7 + …. + (2n – 1) (2n + 1) = $$\frac{n}{3}$$ (4n2 + 6n – 1)

Step I:
Put n = 1
L.H.S. = 1.3 = 3
R.H.S. = $$\frac{1}{3}$$ [4(1)2 + 6(1) – 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is trae for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.3 + 3.5 + 5.7 +….+ (2k – 1)(2k + 1) = $$\frac{k}{3}$$ (4k2 + 6k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.3 + 3.5 + 5.7 +… to n terms = $$\frac{n}{3}$$ (4n2 + 6n – 1) for all n ∈ N.

Question 8.
$$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}$$
Solution:
Let P(n) ≡ $$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}$$, for all n ∈ N.

Step I:
Put n = 1
L.H.S. = $$\frac{1}{1.3}=\frac{1}{3}$$
R.H.S. = $$\frac{1}{2(1)+1}=\frac{1}{3}$$
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ $$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}=\frac{k}{2 k+1}$$ …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ $$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}$$, for all n ∈ N.

Question 9.
$$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}$$
Solution:
Let P(n) ≡ $$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}$$, for all n ∈ N.
But first factor in each term of the denominator,
i.e., 3, 5, 7, ….. are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1) 2 = (2n + 1)
Also, second factor in each term of the denominator,
i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.
∴ nth term = a + (n – 1) d = 5 + (n – 1) 2 = (2n + 3)
∴ nth term, tn = $$\frac{1}{(2 n+1)(2 n+3)}$$
P(n) ≡ $$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}$$ = $$\frac{n}{3(2 n+3)}$$

Step I:
Put n = 1
L.H.S. = $$\frac{1}{3.5}=\frac{1}{15}$$
R.H.S. = $$\frac{1}{3[2(1)+3]}=\frac{1}{3(2+3)}=\frac{1}{15}$$
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ $$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}$$ = $$\frac{k}{3(2 k+3)}$$ ….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ $$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}$$, for all n ∈ N.

Question 10.
(23n – 1) is divisible by 7.
Solution:
(23n – 1) is divisible by 7 if and only if (23n – 1) is a multiple of 7.
Let P(n) ≡ (23n – 1) = 7m, where m ∈ N.

Step I:
Put n = 1
∴ 23n – 1 = 23(1) – 1 = 23 – 1 = 8 – 1 = 7
∴ (23n – 1) is a multiple of 7.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
i.e., 23k – 1 is a multiple of 7.
∴ 23k – 1 = 7a, where a ∈ N
∴ 23k = 7a + 1 ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
23(k+1) – 1 = 7b, where b ∈ N.
∴ P(k + 1) = 23(k+1) – 1
= 23k+3 – 1
= 23k . (23) – 1
= (7a + 1)8 – 1 …..[From (i)]
= 56a + 8 – 1
= 56a + 7
= 7(8a + 1)
7b, where b = (8a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (24n – 1) is divisible by 7, for all n ∈ N.

Question 11.
(24n – 1) is divisible by 15.
Solution:
(24n – 1) is divisible by 15 if and only if (24n – 1) is a multiple of 15.
Let P(n) ≡ (24n – 1) = 15m, where m ∈ N.

Step I:
Put n = 1
∴ 24(1) – 1 = 16 – 1 = 15
∴ (24n – 1) is a multiple of 15.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 24k – 1 = 15a, where a ∈ N
∴ 24k = 15a + 1 …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
∴ 24(k+1) – 1 = 15b, where b ∈ N
∴ P(k + 1) = 24(k+1) – 1 = 24k+4 – 1
= 24k . 24 – 1
= 16 . (24k) – 1
= 16(15a + 1) – 1 …..[From (i)]
= 240a + 16 – 1
= 240a + 15
= 15(16a + 1)
= 15b, where b = (16a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (24n – 1) is divisible by 15, for all n ∈ N.

Question 12.
3n – 2n – 1 is divisible by 4.
Solution:
(3n – 2n – 1) is divisible by 4 if and only if (3n – 2n – 1) is a multiple of 4.
Let P(n) ≡ (3n – 2n – 1) = 4m, where m ∈ N.

Step I:
Put n = 1
∴ (3n – 2n – 1) = 3(1) – 2(1) – 1 = 0 = 4(0)
∴ (3n – 2n – 1) is a multiple of 4.
∴ P(n) is tme for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3k – 2k – 1 = 4a, where a ∈ N
∴ 3k = 4a + 2k + 1 ….(i)

Step III:
We have to prove that P(n) is tme for n = k + 1,
i.e., to prove that
3(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N
P(k + 1) = 3k+1 – 2(k + 1) – 1
= 3k . 3 – 2k – 2 – 1
= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]
= 12a + 6k + 3 – 2k – 3
= 12a + 4k
= 4(3a + k)
= 4b, where b = (3a + k) ∈ N
∴ P(n) is tme for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is tme for all n ∈ N.
∴ 3n – 2n – 1 is divisible by 4, for all n ∈ N.

Question 13.
5 + 52 + 53 + ….. + 5n = $$\frac{5}{4}$$ (5n – 1)
Solution:
Let P(n) ≡ 5 + 52 + 53 +…..+ 5n = $$\frac{5}{4}$$ (5n – 1), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 5
R.H.S. = $$\frac{5}{4}$$ (51 – 1) = 5
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 5 + 52 + 53 + ….. + 5k = $$\frac{5}{4}$$ (5k – 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 5 + 52 + 53 + … + 5n = $$\frac{5}{4}$$ (5n – 1), for all n ∈ N.

Question 14.
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
Solution:
Let P(n) ≡ (cos θ + i sin θ)n = cos nθ + i sin nθ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = (cos θ + i sin θ)1 = cos θ + i sin θ
R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ (cos θ + i sin θ)k = cos kθ + i sin kθ …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
(cos θ + i sin θ)k+1 = cos (k + 1)θ + i sin (k + 1)θ
L.H.S. = (cos θ + i sin θ)k+1
= (cos θ + i sin θ)k . (cos θ + i sin θ)
= (cos kθ + i sin kθ) . (cos θ + i sin θ) ……[From (i)]
= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i2 = -1]
= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)
= cos(kθ + θ) + i sin(kθ + θ)
= cos(k + 1) θ + i sin (k + 1) θ
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (cos θ + i sin θ)n = cos (nθ) + i sin (nθ), for all n ∈ N.

Question 15.
Given that tn+1 = 5 tn+4, t1 = 4, prove by method of induction that tn = 5n – 1.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation tn+1 = 5 tn+4, t1 = 4 and R.H.S. a general statement tn = 5n – 1.
Step I:
Put n = 1
L.H.S. = 4
R.H.S. = 51 – 1 = 4
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S. = t2 = 5t1 + 4 = 24
R.H.S. = t2 = 52 – 1 = 24
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ tk+1 = 5 tk+4 and tk = 5k – 1

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that tk+1 = 5k+1 – 1
Since tk+1 = 5 tk+4 and tk = 5k – 1 …..[From Step II]
tk+1 = 5 (5k – 1) + 4 = 5k+1 – 1
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ tn = 5n – 1, for all n ∈ N.

Question 16.
Prove by method of induction
$$\left(\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right)^{n}=\left(\begin{array}{cc} 1 & 2 n \\ 0 & 1 \end{array}\right) \forall n \in N$$
Solution:

## Maharashtra Board 11th Maths Solutions Chapter 3 Permutations and Combination Miscellaneous Exercise 3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Permutations and Combination Miscellaneous Exercise 3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Miscellaneous Exercise 3

(I) Select the correct answer from the given alternatives.

Question 1.
A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is
(A) 5
(B) 3
(C) 8
(D) 15
(C) 8
Hint:
Number of ways to select one course from available 8 courses
(i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

Question 2.
A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is
(A) 21
(B) 4
(C) 42
(D) 10
(A) 21
Hint:
Number of ways to select one morning and one evening course = 7C1 × 3C1 = 21

Question 3.
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together?
(A) 3! 8!
(B) 3! 4! 8! 4!
(C) 4! 4!
(D) 8! 4! 4!
(B) 3! 4! 8! 4!
Hint:
8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways.
Three groups of Indians, Americans and Englishmen can be permuted in 3! ways.
Required number = 3! × 8! × 4! × 4!

Question 4.
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
(A) 9 × 8!
(B) 8 × 8!
(C) 9 × 9!
(D) 8 × 9!
(D) 8 × 9!
Hint:
Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in 9P2 ways.
Required number = 8! 9P2
= 8! × 9 × 8
= 9! × 8

Question 5.
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
(A) 12
(B) 288
(C) 144
(D) 256
(C) 144
Hint:
B G B G B G B
4 boys take their seats in 4! ways.
3 girls take their seats in 3! ways.
Required number = 4! × 3!
= 24 × 6
= 144

Question 6.
Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
(A) 16
(B) 56
(C) 24
(D) 8
(B) 56
Hint:
A triangle is obtained by joining three vertices.
Number of ways of selecting 3 vertices out of 8 vertices = 8C3
= $$\frac{8 \times 7 \times 6}{1 \times 2 \times 3}$$
= 56

Question 7.
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
(A) 320
(B) 750
(C) 40
(D) 11340
(D) 11340
Hint:
Number of ways to choose 8 questions from Part A and 5 from Part B = 10C8 × 10C5
= 10C2 × 10C5
= 45 × 252
= 11340

Question 8.
There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers is equal to:
(A) 2! × 7!
(B) 2! × 8!
(C) 3! × 7!
(D) 3! × 8!
(B) 2! × 8!
Hint:
Select a person from 8 people (i.e., the people excluding two brothers).
This is done in 8 ways.
2 brothers sit adjacent to the selected person on two sides, they may interchange their seats.
Remaining 7 people sit in 7! ways
Required number = 8 × 2 × 7! = 2! × 8!

Question 9.
The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is
(A) 80
(B) 60
(C) 40
(D) 100
(C) 40
Hint:
Arrange B, A, A, A in $$\frac{4 !}{3 !}$$ ways.
These four letters create 5 gaps in which 2 N are to be filled, this can be done in 5C2 ways, we do not permute those 2N as they are identical.
∴ Required number = $$\frac{4 !}{3 !}$$ × 5C2 = 40

Question 10.
The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is
(A) 840
(B) 600
(C) 720
(D) 480
(D) 480
Hint:
5 males take their seats in 4! ways, creating 5 gaps.
In these 5 gaps, 2 females are to be seated.
∴ The number of ways to do this = 5C2 × 2!
Required number = 4! × 5C2 × 2! = 480

Question 1.
Find the value of r if 56Pr+2 : 54Pr-1 = 30800 : 1.
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
∴ Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R.
Number of words that can be formed by using all these letters = 6! = 720
When a word starts with ‘A’,
‘A’ can be arranged in 1 way and the remaining 5 letters can be arranged among themselves in 5! ways.
The number of words starting with A = 5!
∴ Similarly,
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
Number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
The Capital English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
There are 11 symmetric letters.
∴ Number of 3 Letter passwords = 11P3
= 11 × 10 × 9
= 990

Question 5.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also, among the given numbers 2 is repeated twice and 3 is repeated thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= $$\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}$$
= $$\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}$$
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360

Question 6.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:
Ten students are to be selected for a project from a class of 30 students.
Case I:
If 4 students join the project, then from remaining 26 students, rest of the 6 students are to be selected.
Which can be done in 26C6
= $$\frac{26 !}{6 !(26-6) !}$$
= $$\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 !}{6 ! \times 20 !}$$
= 230230 ways.

Case II:
If 4 students does not join the project, then from remaining 26 students, all the 10 students are to be selected.
Which can be done in 26C10
= $$\frac{26 !}{10 !(26-10) !}$$
= $$\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 !}{10 ! \times 16 !}$$
= 5311735 ways.
∴ Required number of selections = 26C6 + 26C10
= 230230 + 5311735
= 5541965

Question 7.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:
There are 7 books of student’s interest, but he can borrow only three books.
He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed.
Consider the following table:

Required number of selections = 5 + 10 = 15

Question 8.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
Solution:
First we can select 7 objects out of 30 for the first group in 30C7 ways.
Now there are 23 objects left out of which we can select 10 objects for the second group in 23C10 ways.
Remaining 13 objects can be selected for the third group in 5C5 ways.
∴ Required number of ways = 30C7 × 23C10 × 13C13
= $$\frac{30 !}{23 ! 7 !} \times \frac{23 !}{10 ! 13 !} \times 1$$
= $$\frac{30 !}{7 ! 10 ! 13 !}$$

Question 9.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 27 = 128
This number includes one case when the student passes in all subjects.
Required number of ways = 128 – 1 = 127

Question 10.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:
Nine friends decide to go for a picnic in two groups and there must be at least 3 friends in each group.
Consider the following table:

Question 11.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 212
This number includes one case in when all 12 lamps are OFF.
∴ Required Number of ways = 212 – 1 = 4095

Question 12.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
A quadratic equation is to be formed using numbers 0, 2, 4, 5 as coefficients and a coefficient can be repeated.
Let the quadratic equation be ax2 + bx + c = 0, a ≠ 0
Consider the following table:

Number of quadratic equations can be formed = 3 × 4 × 4 = 48

Question 13.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
There are total of 10 digits.
Let the telephone number be 45abcd.
There are 8 digits left for the choice of a, b, c, d as repetition is not allowed.
Consider the following table:

∴ Required number of numbers formed = 8 × 7 × 6 × 5 = 1680

Question 14.
A question paper has 6 questions. How many ways does a student have to answer if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 questions.
Number of outcomes = 26
This number includes one case when the student solves NONE of the questions.
∴ Required number of ways = 26 – 1 = 64 – 1 = 63

Question 15.
Find the number of ways of dividing 20 objects into three groups of sizes 8, 7, and 5.
Solution:
First we can select 8 objects our of 20 for the first group in 20C8 ways.
Now there are 12 objects left out of which we can select 7 objects for the second group in 12C7 ways.
Remaining 5 objects can be selected for the third group in 5C5 ways.
∴ Required number of ways = 20C8 × 12C7 × 5C5
= $$\frac{20 !}{8 ! 12 !} \times \frac{12 !}{7 ! 5 !} \times 1$$
= $$\frac{20 !}{8 ! 7 ! 5 !}$$

Question 16.
There are 4 doctors and 8 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 4 doctors and 8 lawyers in a panel.
A team of 6 with at least one doctor is to be formed.
We count the number by the INDIRECT method of counting.
Number of ways to select a team of 6 people = 12C6
Number of teams with No doctor in any team = 8C6
∴ Required number of ways = 12C68C6
= 924 – 28
= 896

Question 17.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms formed.
Solution:
The first set has 4 parallel lines and another set has 5 parallel lines.
To form a parallelogram, we need 2 lines from each set.
∴ Required number of distinct parallelograms formed = 4C2 × 5C2
= 6 × 10
= 60

Question 18.
There are 12 distinct points A, B, C, …, L, in order, on a circle. Lines are drawn passing through each pair of points.
(i) How many lines are there in total?
(ii) How many lines pass through D?
(iii) How many triangles are determined by lines?
(iv) How many triangles have on vertex C?
Solution:
(i) We need two points to draw a line.
∴ Total number of lines = 12C2 = 66

(ii) Lines are drawn passing through each pair of points.
∴ Lines from point D will pass through all the remaining 11 points.
∴ 11 lines pass through D.

(iii) We need three points to draw a triangle.
∴ Number of triangles = 12C3 = 220

(iv) To get the triangles with one vertex as C,
we need two vertices from the remaining 11 vertices.
∴ Number of triangles with vertex at C = 11C2
= $$\frac{11 \times 10}{2}$$
= 55

## Maharashtra Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Permutations and Combination Ex 3.6 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6

Question 1.
Find the value of
(a) 15C4
Solution:

(b) 80C2
Solution:

(c) 15C4 + 15C5
Solution:

(d) 20C1619C16
Solution:

Question 2.
Find n if
(a) 6P2 = n(6C2)
Solution:

(b) 2nC3 : nC2 = 52 : 3
Solution:

(c) nCn-3 = 84
Solution:

Question 3.
Find r if 14C2r : 10C2r-4 = 143 : 10.
Solution:

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if,
(a) nPr = 720 and nCn-r = 120
Solution:

(b) nCr-1 : nCr : nCr+1 = 20 : 35 : 42
Solution:

Question 5.
If nPr = 1814400 and nCr = 45, find n+4Cr+3.
Solution:

Question 6.
If nCr-1 = 6435, nCr = 5005, nCr+1 = 3003, find rC5.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in 6C3 ways.
3 reen balls can be selected from 8 green balls in 8C3 ways.
3 blue balls can be selected from 7 blue balls in 7C3 ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in 6C3 ways.
2 girls can be selected from 4 girls in 4C2 ways.
∴ Number of ways the team can be selected

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = nC2
Given 66 handshakes were exchanged.
66 = nC2
66 = $$\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}$$
66 × 2 = $$\frac{n(n-1)(n-2) !}{(n-2) !}$$
132 = n (n – 1)
n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
(iv) n = 8
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw nC2 lines including sides.
∴ Number of diagonals in n sided polygon = nC2 – n (n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel, they intersect at a point.
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = 20C2
= $$\frac{20 !}{2 ! 18 !}$$
= $$\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}$$
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinear
Solution:
There are 10 points on a plane.
(a) When no three of them are collinear.
A line is obtained by joining 2 points.
∴ Number of lines passing through these points = 10C2
= $$\frac{10 !}{2 ! 8 !}$$
= $$\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}$$
= 5 × 9
= 45

(b) When 4 of them are collinear.
If no three points are collinear, we get a total of 10C2 = 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of 4C2 lines.
4C2 – 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (4C2 – 1)
= 45 – $$\frac{4 !}{2 ! 2 !}$$ + 1
= 45 – $$\frac{4 \times 3 \times 2 !}{2 \times 2 !}$$ + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(a) no three points are collinear
(b) four points are collinear
Solution:
There are 12 points on the plane.
(a) When no three of them are collinear.
A triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = 12C3
= $$\frac{12 !}{3 ! 9 !}$$
= $$\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}$$
= 220

(b) When 4 of these points are collinear.
If no three points are collinear, total we get 12C3 = 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
4C3 extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points = 12C34C3
= 220 – $$\frac{4 !}{3 ! \times 1 !}$$
= 220 – $$\frac{4 \times 3 !}{3 !}$$
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
There are 8 consonants and 3 vowels.
From 8 consonants, 4 can be selected in 8C4
= $$\frac{8 !}{4 ! 4 !}$$
= $$\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}$$
= 70 ways.
From 3 vowels, 2 can be selected in 3C2
= $$\frac{3 !}{2 ! 1 !}$$
= $$\frac{3 \times 2 !}{2 !}$$
= 3 ways.
Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in 6P6 = 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Question 16.
Find n if,
(i) nC8 = nC12
Solution:
nC8 = nC12
If nCx = nCy, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

(ii) 23C3n = 23C2n+3
Solution:
23C3n = 23C2n+3
If nCx = nCy, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

(iii) 21C6n = $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$
Solution:
21C6n = $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$
If nCx = nCy, then either x = y or x = n – y
∴ 6n = n2 + 5 or 6n = 21 – (n2 + 5)
∴ n2 – 6n + 5 = 0 or 6n = 21 – n2 – 5
∴ n2 – 6n + 5 = 0 or n2 + 6n – 16 = 0
If n2 – 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n2 + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n2 + 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2

Check:
n = 2
∴ n2 + 5 = 22 + 5 = 9
21C6n = 21C12
and $${ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}$$ = 21C9
nCr = nCn-r
21C12 = 21C9
∴ n = 2 is a right answer.

(iv) 2nCr-1 = 2nCr+1
Solution:
2nCr-1 = 2nCr+1
If nCx = nCy, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Check:
2nCr-1 = 2nCn-1
and 2nCr+1 = 2nCn+1
using nCr = nCn-r, we have
2nCn+1 = 2nC2n-(n+1) = 2nCn-1
2nCr-1 = 2nCr+1

(v) nCn-2 = 15
Solution:
nCn-2 = 15
nC2 = 15 …..[∵ nCr = nCn-r]
∴ $$\frac{n !}{(n-2) ! 2 !}=15$$
∴ $$\frac{n(n-1)(n-2) !}{(n-2) ! \times 2 \times 1}=15$$
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6

Question 17.
Find x if nPr = x nCr.
Solution:

Question 18.
Find r if 11C4 + 11C5 + 12C6 + 13C7 = 14Cr.
Solution:

Question 19.
Find the value of $$\sum_{r=1}^{4}{ }^{(21-r)} \mathrm{C}_{4}$$.
Solution:

Question 20.
Find the differences between the greatest values in the following:
(a) 14Cr and 12Cr
Solution:
Greatest value of 14Cr.
Here, n = 14, which is even.
Greatest value of nCr occurs at r = $$\frac{n}{2}$$ if n is even.
∴ r = $$\frac{n}{2}$$
∴ r = $$\frac{14}{2}$$ = 7

∴ Difference between the greatest values of 14Cr and 12Cr = 14Cr12Cr
= 3432 – 924
= 2508

(b) 13Cr and 8Cr
Solution:
Greatest value of 13Cr.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = $$\frac{n-1}{2}$$ if n is odd.
∴ r = $$\frac{\mathrm{n}-1}{2}$$
∴ r = $$\frac{13-1}{2}$$ = 6

∴ Difference between the greatest values of 13Cr and 8Cr = 13Cr8Cr
= 1716 – 70
= 1646

(c) 15Cr and 11Cr
Solution:
Greatest value of 15Cr.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = $$\frac{n-1}{2}$$ if n is odd.
∴ r = $$\frac{n-1}{2}$$
∴ r = $$\frac{15-1}{2}$$ = 7

∴ Difference between the greatest values of 15Cr and 11Cr = 15Cr11Cr
= 6435 – 462
= 5973

Question 21.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16

Question 22.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consist of at least 3 women.
Consider the following table:

∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 23.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.
Consider the following table:

∴ Number of committees with at least 5 women
= 14112 + 14700 + 6720 + 1260 + 81
= 36873

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= 18C10 – 36873
= 18C8 – 36873
= 43758 – 36873
= 6885

Question 24.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 25.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected
= 45045 + 6006
= 51051

Question 26.
Five students are selected from 11. How many ways can these students be selected if
(a) two specified students are selected?
(b) two specified students are not selected?
Solution:
5 students are to be selected from 11 students.
(a) When 2 specified students are included,
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = 9C3
= $$\frac{9 !}{3 ! \times 6 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}$$
= 84
∴ Selection of students is done in 84 ways when 2 specified students are included.

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students = 9C5
= $$\frac{9 !}{5 ! 4 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}$$
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included.

## Maharashtra Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.5

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Permutations and Combination Ex 3.5 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.5

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways.
∴ 8 friends can sit around a table in 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.

Question 2.
A party has 20 participants. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants are seated on either side of the host.
The host takes the chair in 1 way.
These 2 persons can sit on either side of the host in 2! ways.
Once the host occupies his chair, it is not circular permutation more.
The remaining 18 people occupy their chairs in 18! ways.
∴ A total number of arrangements possible if two particular participants are seated on either side of the host = 2! × 18! = 2 × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are (a) always together. (b) never together.
Solution:
(a) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit. They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total of 23) which can be done in (23 – 1)! = 22! ways.
∴ Required number of arrangements = 2! × 22!

(b) When 2 specified delegates are never together then, other 22 delegates can be participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates so that they are never together.
∴ Two specified delegates can be arranged in 22P2 ways.
∴ Required number of arrangements = 22P2 × 21!
= $$\frac{22 !}{(22-2) !} \times 21 !$$
= $$\frac{22 !}{20 !}$$ × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in(15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e., clockwise and anticlockwise) coincide.

∴ Required number of arrangements = $$\frac{14 !}{2}$$

Question 5.
A committee of 10 members sits around a table. Find the number of arrangements that have the President and the Vice-president together.
Solution:
A committee of 10 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 8 members of the committee is to be arranged around a table, which can be done in (9 – 1)! = 8! ways.
∴ Required number of arrangements = 8! × 2! = 2 × 8!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated (a) between the two women. (b) between two men.
Solution:
5 men, 2 women, and a child sit around a table.
(a) When the child is seated between two women.
5 men, 2 women, and a child are to be seated around a round table such that the child is seated between two women.
∴ the two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Now, this one unit is to be arranged with the remaining 5 men,
i.e., a total of 6 units are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 5! × 2!
= 120 × 2
= 240

(b) Two men can be selected from 5 men in
5C2 = $$\frac{5 !}{2 !(5-2) !}=\frac{5 \times 4 \times 3 !}{2 \times 3 !}$$ = 10 ways.
Also, these two men can sit on either side of the child in 2! ways.
Let us take two men and a child as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 2 women,
i.e., a total of 6 units (3 + 2 + 1) are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 10 × 2! × 5!
= 10 × 2 × 120
= 2400

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
No two women should sit together.
There are 8 gaps created by 8 men’s seats.
∴ Women can be seated in 8 gaps in 8P6 ways.
∴ Required number of arrangements = 7! × 8P6

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:
2 women (who wish to sit together) can be selected from 3 in
3C2 = $$\frac{3 !}{2 !(3-2) !}=\frac{3 \times 2 !}{2 ! \times 1 !}$$ = 3 ways.
Also, these two women can sit together in 2! ways.
Let us take two women as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 1 woman,
i.e., a total of 5 units are to be arranged around a round table, which can be done in (5 – 1)! = 4! ways.
∴ Required number of arrangements = 3 × 2! × 4!
= 3 × 2 × 24
= 144

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in $$\frac{9 !}{4 !}$$ ways.
∴ Required number of arrangements = $$\frac{9 !}{4 !}$$

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side,
the other 13 persons can be arranged around the table in (13 – 1)! = 12! ways.
The two persons who are not sitting side by side may take 13 positions created by 3 persons in 13P2 ways.
∴ Required number of arrangements = 12! × 13P2
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

## Maharashtra Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.4

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Permutations and Combination Ex 3.4 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.4

Question 1.
Find the number of permutations of letters in each of the following words.
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
(v) BAL BHARATI
Solution:
(i) There are 5 distinct letters in the word DIVYA.
∴ Number of permutations of the letters of the word DIVYA = 5! = 120

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = $$\frac{9 !}{3 !}$$
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.
∴ Number of permutations of the letters of the word REPRESENT = $$\frac{9 !}{3 ! 2 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}$$
= 30240

(iv) There are 7 distinct letters in the word COMBINE.
∴ Number of permutations of the letters of the word COMBINE = 7! = 5040

(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word BALBHARATI = $$\frac{10 !}{2 ! 3 !}$$
= $$\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 2}$$
= 302400

Question 2.
You have 2 identical books on English, 3 identical books on Hindi, and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on mathematics are identical.
∴ Total number of arrangements possible = $$\frac{9 !}{2 ! 3 ! 4 !}$$
= $$\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}$$
= 9 × 4 × 7 × 5
= 1260

Question 3.
A coin is tossed 8 times. In how many ways can we obtain (a) 4 heads and 4 tails? (b) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(a) 4 heads and 4 tails are to be obtained.
∴ Number of ways it can be obtained = $$\frac{8 !}{4 ! 4 !}$$
= $$\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}$$
= 70

(b) At least 6 heads are to be obtained.
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways it can be obtained = $$\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}$$
= $$\frac{8 \times 7}{2}$$ + 8 + 1
= 28 + 8 + 1
= 37

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = $$\frac{13 !}{5 ! 4 ! 4 !}$$

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times.
∴ Required number of arrangements = $$\frac{12 !}{2 ! 3 ! 2 !}$$
When all the vowels,
i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.
Let us consider them as one unit.
Number of arrangements of these vowels among themselves = $$\frac{5 !}{3 !}$$ ways.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of such arrangements = $$\frac{8 !}{2 ! 2 !}$$
∴ Required number of arrangements = $$\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}$$

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have (a) letters R and H never together? (b) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of words can be formed = $$\frac{11 !}{4 ! 2 ! 2 !}$$

(a) When letters R and H are never together.
Other than 2R, 2H there are 4A, 1S, 1T, 1M.
These letters can be arranged in $$\frac{7 !}{4 !}$$ ways = 210.
These seven letters create 8 gaps in which 2R, 2H are to be arranged.
Number of ways to do = $$\frac{{ }^{8} \mathrm{P}_{4}}{2 ! 2 !}$$ = 420
Required number of arrangements = 210 × 420 = 88200.

(b) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA, i.e., A, A, A, A.
Let us consider these 4 vowels as one unit, which can be arranged among themselves in $$\frac{4 !}{4 !}$$ = 1 way.
This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.
∴ Total number of arrangements = $$\frac{8 !}{2 ! 2 !}$$

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
To find the number of different words when ‘R’ is taken thrice, ‘S’ is taken twice and ‘T’ is taken twice.
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of words = $$\frac{7 !}{2 ! 2 ! 3 !}$$
= $$\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}$$
= 7 × 6 × 5
= 210

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit.
This unit with the other 4 letters in which ‘M’ repeats twice is to be arranged.
∴ Required number of arrangements = $$\frac{5 !}{2 !}$$
= $$\frac{5 \times 4 \times 3 \times 2 !}{2 !}$$
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, ‘I’ repeat two times each, ‘T’ repeats 3 times.
When the arrangement starts and ends with ‘N’,
other 10 letters can be arranged between two N,
in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Required number of arrangements = $$\frac{10 !}{2 ! 2 ! 3 !}$$

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements do not have the two R’s and two A’s together?
Solution:
There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each.
∴ Number of ways to arrange the letters of word ARRANGE = $$\frac{7 !}{2 ! 2 !}$$ = 1260
Consider the words in which 2A are together and 2R are together.
Let us consider 2A as one unit and 2R as one unit.
These two units with remaining 3 letters can be arranged in = $$\frac{5 !}{2 ! 2 !}$$ = 30 ways.
Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = $$\frac{5 !}{2 !}+\frac{5 !}{2 !}$$ = 5! = 120
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = $$\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}$$ = 60
Required number of numbers = 120 + 60 = 180

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits, i.e. 3, 5, 7, 9.
∴ They can be arranged at 4 odd places among themselves in 4! = 24 ways.
There are 3 even digits, i.e. 4, 6, 8.
∴ They can be arranged at 3 even places among themselves in 3! = 6 ways.
∴ Required number of numbers formed = 24 × 6 = 144

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 4?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Required number of numbers formed = $$\frac{6 !}{2 !}$$
= $$\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}$$
= 360
A 6-digit number is to be formed using the same digits that are divisible by 4.
For a number to be divisible by 4, the last two digits should be divisible by 4,
i.e. 24, 52, 56, 64, 92 or 96.
Case I: When the last two digits are 24, 52, 56 or 64.
As the digit 9 repeats twice in the remaining four numbers, the number of arrangements = $$\frac{4 !}{2 !}$$ = 12
∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48.
Case II: When the last two digits are 92 or 96.
As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24
∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48.
∴ Required number of numbers framed = 48 + 48 = 96

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N are repeated twice.
Number of different words that can be formed using the letters of the word INDIAN = $$\frac{6 !}{2 ! 2 !}$$
= $$\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}$$
= 180
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in $$\frac{5 !}{2 !}$$
= $$\frac{5 \times 4 \times 3 \times 2 !}{2 !}$$
= 60 ways.
∴ 2 N can be arranged in $$\frac{2 !}{2 !}$$ = 1 way.
∴ Required number of words = 60 × 1 = 60

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if (a) the two O’s are never together. (b) consonants and vowels occupy alternate positions.
Solution:
There are 7 letters in the words PLATOON in which ‘O’ repeat 2 times.
(a) When the two O’s are never together.
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
Two O’s can take their places in 6P2 ways.
But ‘O’ repeats 2 times.
∴ Two O’s can be arranged in $$\frac{{ }^{6} \mathrm{P}_{2}}{2 !}$$
= $$\frac{\frac{6 !}{(6-2) !}}{2 !}$$
= $$\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}$$
= 3 × 5
= 15 ways
∴ Required number of arrangements = 120 × 15 = 1800

(b) When consonants and vowels occupy alternate positions.
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places, consonants occur and at even places, vowels occur.
4 consonants can be arranged among themselves in 4! ways.
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in $$\frac{3 !}{2 !}$$ ways.
Now, vowels and consonants should occupy alternate positions.
∴ Required number of arrangements = 4! × $$\frac{3 !}{2 !}$$
= 4 × 3 × 2 × $$\frac{3 \times 2 !}{2 !}$$
= 72