Exercise

Permutation and Combination

Question 1

• Given A = {1,2,3,4,5,6,7 }. How many three digit numbers can be generated without repeating any digit?

Answer

• 7 x 6 x 5 = 210

Question 2

• There are 8 persons in an office. A group consisting of 5 persons has to be formed. In how many ways can the group be formed?

Answer

• Number of ways in which the group can be formed = 8C5

Question 3

• In how many ways can a team of 5 persons can be formed out of a total of 10 persons such that two particular persons should be included in each team?

Answer

• Two particular persons should be included in each team

i.e., we have to select 5-2 = 3 persons from 10-2 = 8 persons

Hence, the required number of ways = 8C3

= 8×7×63×2×1

= 8×7= 56

Question 4

• There are 10 book at the shelf. 2 of the book have the same title. In how many ways the books can be arranged on a shelf such that a same title books should always be together?

Answer• We have a total of 10 books.

Given that a particular pair of books should always be together. Hence, just tie thesetwo books together and consider as a single book.

Hence we can take total number of books as 9. These 9 books can be arranged in 9P9 = 9! Ways

We had tied two books together. These books can be arranged among themselves in 2P2 = 2! Ways

Hence, the required number of ways = 9! × 2!

Question 5

• There are 3 books each of 5 volumes and two books each of two volumes. In how many ways can these books be arranged in a shelf so that the volumes of the same book should remain together?

Answer• 1 book : 5 volume

1 book : 5 volume1 book : 2 volume1 book : 2 volume

Given that volumes of the same book should remain together. Hence, just tie thesame volume books together and consider as a single book. Hence we can take totalnumber of books as 4. These 4 books can be arranged in 4P4 = 4! Ways

The 5 volumes of the 1st book can be arranged among themselves in 5P5 = 5! Ways

The 5 volumes of the 2st book can be arranged among themselves in 5P5 = 5! Ways

The 2 volumes of the 3rd book can be arranged among themselvesin 2P2 = 2! Ways

The 2 volumes of the 4th book can be arranged among themselves in 2P2 = 2! Ways

Hence total number of ways = 4! × 5! × 5! × 2! × 2!

Question 6

• In how many ways can 15 persons be arranged in a row such that 4 particular persons should always be together?

Answer• Given that three particular persons should always be together.

Hence, just group these three persons together and consider as a single person.

Hence we can take total number of persons as 11. These 9 persons can be arranged in 9119 = 11! Ways

We had grouped three persons together. These three persons can be arranged among themselvesin 4P4 = 4! Ways

Hence, the required number of ways = 11! × 4!

Question 7

• How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

Answer

• The first two places can only be filled by 3 and 5 respectively and there is only 1 way of doing this

Given that no digit appears more than once. Hence we have 8 digits remaining(0,1,2,4,6,7,8,9)So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways

Total number of ways = 8P4 = 8 x 7 x 6 x 5 = 1680

Question 8

• 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?

Answer

• He can go in any bus out of the 25 buses. Hence He can go in 25 ways.

Since he can not come back in the same bus that he used for travelling,He can return in 24 ways.

Total number of ways = 25 x 24 = 600

Question 9

• A box contains 6 blue, 4 green and 3 pink balls. Three balls are drawn at random. Find out the number of ways of selecting the 2 red, 2 white and 1 blue balls.

Answer

• 2 blue ball can be selected in 6C2 ways2 green ball can be selected in 4C2 ways1 pink ball can be selected in 3C1 ways

= 6C2 x 4C2 x 3C1

Question 10

• Seven women and nine men are on the mathematics club at a school. How many ways are there to select a committee of FIVE (5) members of the club if at least one woman must be on the committee?

Answer

• 7C1 x 9C4 + 7C2 x 9C3 + 7C3 x 9C2 + 7C4 x 9C1 + 7C5

Question 11

• Seven women and nine men are on the mathematics club at a school. How many ways are there to select a committee of FIVE (5) members of the club if at least one woman and at least one man must be on the committee?

Answer

• 7C1 x 9C4 + 7C2 x 9C3 + 7C3 x 9C2 + 7C4 x 9C1