Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.1

Question 1.

A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?

Solution:

There are 30 boys and 20 girls in a class.

The teacher wants to select a class monitor from these boys and girls.

A boy can be selected in 30 ways and a girl can be selected in 20 ways.

∴ By using the fundamental principle of addition,

in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.

Question 2.

In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?

Solution:

(i) Since there are 30 boys in the class

∴ A boy monitor can be selected in 30 ways.

(ii) Since there are 20 girls in the class

∴ A girl monitor can be selected in 20 ways.

Question 3.

A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?

Solution:

A signal is generated from 2 flags and there are 4 flags of different colours available.

∴ 1st flag can be any one of the available 4 flags.

∴ It can be selected in 4 ways.

Now, 2nd flag is to be selected for which 3 flags are available for a different signal.

∴ 2nd flag can be anyone from these 3 flags.

∴ It can be selected in 3 ways.

∴ By using the fundamental principle of multiplication,

Total number of ways in which a signal can be generated = 4 × 3 = 12

∴ 12 different signals can be generated.

Question 4.

How many two-letter words can be formed using letters from the word SPACE when repetition of letters

(i) is allowed

(ii) is not allowed

Solution:

A two-letter word is to be formed out of the letters of the word SPACE.

(i) When repetition of the letters is allowed

1st letter can be selected in 5 ways

2nd letter can be selected in 5 ways

∴ By using the fundamental principle of multiplication,

total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed

1st letter can be selected in 5 ways

2nd letter can be selected in 4 ways

∴ By using the fundamental principle of multiplication,

total number of 2-letter words = 5 × 4 = 20

Question 5.

How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits

(i) are allowed

(ii) are not allowed

Solution:

The three-digit number is to be formed from the digits 0, 1, 3, 5, 6

(i) When repetition of digits is allowed:

100’s place digit should be a non-zero number.

Hence, it can be anyone from digits 1, 3, 5, 6

∴ 100’s place digit can be selected in 4 ways.

0 can appear in 10’s and unit’s place and digits can be repeated.

∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.

∴ By using the fundamental principle of multiplication,

the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed:

100’s place digit should be a non-zero number.

Hence, it can be anyone from digits 1, 3, 5, 6

∴ 100’s place digit can be selected in 4 ways

0 can appear in 10’s and unit’s place and digits can’t be repeated.

∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways

∴ By using the fundamental principle of multiplication,

total number of three-digit numbers = 4 × 4 × 3 = 48

Question 6.

How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?

Solution:

A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.

∴ The unit’s place digit can be selected in 5 ways.

10’s place digit can be selected in 5 ways.

100’s place digit can be selected in 5 ways.

∴ By using fundamental principle of multiplication,

the total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 7.

A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.

Solution:

A letter lock has 3 rings, each ring containing 5 different letters.

∴ A letter from each ring can be selected in 5 ways.

∴ By using fundamental principle of multiplication,

the total number of trials that can be made = 5 × 5 × 5 = 125

Out of these 124 wrong attempts are made and in the 125th attempt,

the lock gets opened, for a maximum number of trials.

∴ A maximum number of trials required to open the lock is 125.

Question 8.

In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?

Solution:

For a set of 5 true/false questions, each question can be answered in 2 ways.

∴ By using fundamental principle of multiplication,

the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32

Since no student has written all the correct answers.

∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31

Also, no student has given the same sequence of answers.

∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31

Question 9.

How many numbers between 100 and 1000 have 4 in the unit’s place?

Solution:

Numbers between 100 and 1000 are 3-digit numbers.

A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.

Since Unit’s place digit is 4.

∴ it can be selected in 1 way only.

10’s place digit can be selected in 10 ways.

For 3-digit number 100’s place digit should be a non-zero number.

∴ 100’s place digit can be selected in 9 ways.

∴ By using fundamental principle of multiplication,

total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90

Question 10.

How many numbers between 100 and 1000 have the digit 7 exactly once?

Solution:

Numbers between 100 and 1000 are 3-digit numbers.

A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.

When 7 is in the unit’s place

The unit’s place digit is 7.

∴ it can be selected in 1 way only.

10’s place digit can be selected in 9 ways.

100’s place digit can be selected in 8 ways.

∴ total number of numbers which have 7 in the unit’s place = 1 × 9 × 8 = 72

When 7 is in 10’s place

The unit’s place digit can be selected in 9 ways.

10’s place digit is 7

∴ it can be selected in 1 way only.

100’s place digit can be selected in 8 ways.

∴ total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72

When 7 is in 100’s place

The unit’s place digit can be selected in 9 ways.

10’s place digit can be selected in 9 ways.

100’s place digit is 7

∴ it can be selected in 1 way.

∴ total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81

∴ total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.

Question 11.

How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?

Solution:

Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.

∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.

∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.

∴ 1000’s place digit can be selected in 4 ways.

100’s place digit can be selected in 3 ways.

10’s place digit can be selected in 2 ways.

The unit’s place digit can be selected in 1 way.

∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7

= Total number of four-digit numbers formed from the digits 2, 3, 4, 7

= 4 × 3 × 2 × 1

= 24

Question 12.

If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?

Solution:

Case I: Three-digit numbers with 4 occurring in hundred’s place:

100’s place digit can be selected in 1 way.

Ten’s place can be filled by any one of the numbers 2, 3, 5, 6.

∴ 10’s place digit can be selected in 4 ways.

The unit’s place digit can be selected in 3 ways.

∴ total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12

Case II: Three-digit numbers more than 500

100’s place digit can be selected in 2 ways.

10’s place digit can be selected in 4 ways.

Unit’s place digit can be selected in 3 ways.

∴ total number of three digit numbers more than 500 = 2 × 4 × 3 = 24

Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6

Since, repetition of digits is not allowed

∴ total four digit numbers formed = 5 × 4 × 3 × 2 = 120

Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6

Since, repetition of digits is not allowed

∴ total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120

∴ total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276

Question 13.

How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?

Solution:

Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8

Number of such numbers = 3

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8

Ten’s place digit is selected from 2, 5, 7, 8.

∴ Ten’s place digit can be selected in 4 ways.

Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8

∴ The unit’s place digit can be selected in 6 ways.

Using the multiplication principle,

the number of such numbers (repetition allowed) = 4 × 6 = 24

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8

100’s place digit is anyone from 1, 2, 5, 7, 8.

∴ 100’s place digit can be selected in 5 ways.

As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8

∴ 10’s place and unit’s place digits can be selected in 6 ways each.

Using multiplication principle,

the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180

All cases are mutually exclusive and exhaustive.

∴ Required number = 3 + 24 + 180 = 207

Question 14.

A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?

Solution:

A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.

∴ Number of ways to choose gates = 3

Number of ways to choose staircase = 4

∴ By using fundamental principle of multiplication,

number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 15.

How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?

Solution:

For a number to be divisible by 3.

The sum of digits must be divisible by 3.

Given 6 digits are 0, 1,2, 3, 4, 5.

Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.

∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.

Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)

Case I:

3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.

10000’s place digit can be selected in 4 ways (as 0 cannot appear).

As digits are not repeated, 1000’s place digit can be selected in 4 ways.

100’s place digit can be selected in 3 ways.

10’s place digit can be selected in 2 ways.

The unit’s place digit can be selected in 1 way.

∴ Using multiplication theorem,

Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96

Case II:

3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5

10000’s place digit can be selected in 5 ways.

1000’s place digit can be selected in 4 ways.

100’s place digit can be selected in 3 ways.

10’s place digit can be selected in 2 ways.

The unit’s place digit can be selected in 1 way.

Using multiplication theorem,

Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120

Both the cases are mutually exclusive and exhaustive.

∴ Required number = 96 + 120 = 216