Question 1

Find the number of ways, in which 12 different beads can be arranged to form a necklace .

Accepted Answer

Number of arrangements of beads = (12 - 1) !, but it is not mentioned that either it is clockwise or anti - clockwise . So, required number of arrangements = [(12 - 1)! ] / 2 = 11 !/2

Question 2

20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?

Accepted Answer

Total persons on the circular table = 20 guests + 1 host = 21 They can be seated in (21 - 1) ! = 20 ! ways .

Question 3

In how many different ways, 5 boys and 5 girls can sit on a circular table, so that the boys and girls are alternate?

Accepted Answer

After fixing up one boy on the table, the remaining can be arranged in 4 ! ways, but boys and girls have to be alternate. There will be 5 places, one place each between two boys. These 5 place can be filled by 5 girls in 5 ! ways . Hence, by the principle of multiplication, the required number of ways = 4 ! x 5 ! = 2880

Question 4

How many necklaces of 12 beads can be made from 18 beads of various colours?

Accepted Answer

First, we can select 12 beads out of 18 beads in 18C12 ways. Now, these 12 beads can make a necklace in 11! / 2 ways as clockwise and anti-clockwise arrangements are same. So, required number of ways = [ 18C12 .11! ] / 2! = [18C6.11! ] / 2! = [18 x 17 x 16 x 15 x 14 x 13 x 11! ] / [6 x 5 x 4 x 3 x 2 x 1 x 2!] = [17 x 7 x 13! ] / 2! = 119 x 13 !/ 2

Question 5

In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?

Accepted Answer

Total number of ways = 4C3 x 4C2 = [4! / {3! x 1!}] x [4! /{2! x 2!}] = (4 x 4 x 3 x 2 x 1) / (2 x 2) = 4 x 6 = 24

Question 6

In how many ways, a cricket team of 11 players can be made from 15 players, if a particulars player is always chosen?

Accepted Answer

Since, particular player is always chosen. It means that 11 - 1 = 10 players are selected out of the remaining 15 - 1 = 14 players. ∴ Required number of ways = 14C10 = 14 ! / (10! x 4!) = (14 x 13 x 12 x 11) / (4 x 3 x 2 x 1) = 7 x 13 x 11 = 91 x 11 = 1001

Question 7

There is a polygon of 12 sides . How many triangles can be drawn using the vertices of polygon?

Accepted Answer

Required number of triangles = 12C3 = (12 x 11 x 10) / 6 = 220

Question 8

Find the number of diagonals formed in hexagon .

Accepted Answer

Hexagon has 6 sides . ⇒ n = 6 ∴ Required number of diagonals = 10C2 - n = 6C2 - 6 = 6! / [2!(6 - 2)!] - 6 = 6! / (2!4!) - 6 = (6 x 5 x 4!) / (2 x 4!) - 6 = 15 - 6 = 9

Question 9

There are 10 points in a plan, out of which 5 are collinear. Find the number of straight lines formed by joining them.

Accepted Answer

Required number of straight lines =nC2 - mC2 + 1 Here, n = 10, m = 5 = 10C2 - 5C2 + 1 = 45 - 10 + 1 = 36

Question 10

In how many ways, 12 balls can be divided between 2 boys, one receiving 5 and the other 7 balls?

Accepted Answer

Here, order is important, then the number of ways in which 12 different balls can be divided between two boys who receives 5 and 7 balls respectively, is = [12! / 5! 7! ] x 2! = 1584

Question 11

Find the number of combinations that can be formed with 5 oranges, 4 mangoes and 3 bananas, when one fruit of each kind is taken.

Accepted Answer

The required number of combinations, when one fruit of each kind is taken = 5C1 x 4C1 x 3C1 = 5 x 4 x 3 = 60

Question 12

There are 10 questions in a question paper. In how many ways, a student can solve there questions, if he solves one or more question?

Accepted Answer

Required number of ways = 2n - 1 = 210 - 1 = 1024 - 1 = 1023

Question 13

In how many ways, can 15 people be seated around two round tables with seating capacities of 7 and 8 people?

Accepted Answer

Number of ways in which 8 persons can be selected from 15 persons = 15C8 Now, 8 persons can be seated around a circular table in 7! ways. Now, remaining 7 persons can be seated around a circular table in 6! ways. ∴ Required number of ways = 15C8 x 7! x 6!

Question 14

O and A occupying end places.

Accepted Answer

When O and A occupy end places. Then, the three letters (M, E, G) can be arranged themselves by 3! = 6 ways and two letters (O, A) can be arranged among themselves in 2! = 2 ways. ∴ Total number of ways = 6 x 2 = 12

Question 15

E being always in the middle.

Accepted Answer

When E is fixed in the middle, then there are four places left to be filled by four remaining letters O, M, G and A and this can be done in 4! ways. ∴ Total number of ways = 4! = 24

Question 16

Vowels occupying odd places .

Accepted Answer

Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place). ∴ Total number of ways = 3! X 2! = 12

Question 17

Vowels being never together.

Accepted Answer

Total number of words = 5! = 120 combining the vowels at one place(OEA) with remaining 2 letters MG, letters can be arranged in 3! ways. Also, three vowels can be arranged in 3! ways So, when vowels are together, then number of words = 3! x 3! = 36 there4; Required number of ways, when vowels being never together =120 - 36 = 84

Question 18

In how many ways can the letters of the word 'Director' be arranged so that the three vowels are never together?

Accepted Answer

Total number of letters = 8 Number of vowels = 3 and r occurs twice. Total number of arrangements when there is no restriction = 8!/2! When three vowels are together, regarding them as one letter, we have only 5 + 1 = 6 letters These six letters can be arranged in 6!/2! ways, since r occurs twice. But the three vowels can be arranged among themselves in 3! ways. Hence number of arrangement when the three vowels are together = 6! /(2 !x 3!) ∴ Required number = 8!/2! - {6! / (2! x 3!)} = 18,000

Question 19

A mixed doubles tennis game is to be played between two teams ( each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?

Accepted Answer

Married couples : MF MF MF MF ⇒ AB, CD, EF, CD Possible teams : AD CB EB GB AF CF ED GD AH CH EH GF Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams). Teams AD cannot play with AF, AH, ED and GD. The same will apply with all teams, So, number of total matches = 12 x 7 = 84 But every match includes 2 teams, so the actual number of matches = 84/2 = 48

Question 20

There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone number are possible?

Accepted Answer

There is a 7 - digit telephone number but extreme right and extreme left positions are fixed . i. e. , 6 x x x x x 5 ∴ Required number of ways = 8 x 7 x 6 x 5 x 4 = 6720

Permutation and Combination Questions