Question 1

If nP r= 720 nC r than the value r is?

Accepted Answer

We know that, nPr = nCrr! ∵ nPr = 720nCr ⇒ nCr.r! = 720nCr ⇒ r! = 720 ∴ r = 6

Question 2

If S= {2, 3, 4, 5, 7, 9} , then the number of different three-digit number (with all distinct digits) less than 400 that can be formed from S is?

Accepted Answer

The hundred place will be reserved for 3 or 2 5 digits are free to fill rest two places i,e., of tens and unit. Number of required 3 digit number = 2 x 5C2 = 20

Question 3

If 7 points out of 12 are in the same straight line, then the number of triangles formed is?

Accepted Answer

The number of ways of selecting 3 points out of 12 points is 12C3. The number of ways of selecting 3 points out of 7 points, on the same straight line is 7C3, Hence, the number of triangle formed will be 12C3 - 7C3 = 210 - 35 = 185.

Question 4

In how many different ways can the letters of word JUDGE be arranged so that the vowels always come together?

Accepted Answer

Total number of letters = 5 Number of vowels = 2. If we consider both vowel as a one letter then, Required number = 4! 2! = 48.

Question 5

From 4 officers and 8 Jawans in how many ways can 6 be chosen to include at least one officer?

Accepted Answer

We may choose 1 officer and 5 jawans or 2 officers and 4 jawans ......... or 4 officers and 2 jawans. So Required answer = [4C1 x 8C5] + [4C2 x 8C4] + [4C3 x 8C3] + [4C4 x 8C2] = 224 + 420 + 224 + 28 = 896.

Question 6

From a group of 6 men and 4 women a committee of 4 persons is to be formed?

Accepted Answer

Required no. of ways = [4C1 x 6C3] + [4C2 x 6C2] + [4C3 x 6C1] + [4C4] = 80 + 90 + 24 + 1 = 195.

Question 7

The numbers of straight lines can be formed out of 10 points of which 7 are collinear?

Accepted Answer

If there were no three points collinear. We should have 10C2 lines but since 7 points are collinear we must subtract 7C2 lines and add the one corresponding to the line of collinearity of the seven points. Thus, the required number of straight lines . = 10C2 - 7C2 + 1 = 25

Question 8

The number of different permutations of the word 'BANANA' is?

Accepted Answer

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words = 6! / (3! 2!) = 60

Question 9

Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is?

Accepted Answer

Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n. nC2 = 66 ⇒ n(n - 1) / 2! = 66 ⇒ n2 - n - 132 = 0 ⇒ (n - 12) (n + 11) = 0 ∴ n = 12

Question 10

In how many ways can the letter of the word ' civilization' be arranged?

Accepted Answer

There are 12 letters in the world 'civilization' of which four are i's and other are different letters. ∴ Total number of permutations = 12!/4! But one word is civilization itself. ∴ Required number of rearrangements = 12!/4! - 1

Question 11

Letters of the word DIRECTOR are arranged in such a way that all the vowels come together. Find out the total no. of ways for making such arrangement.?

Accepted Answer

Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s Hence no. of arrangement = 6!/2! x 3! = 2160 [3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]

Question 12

How many different letter arrangements can be made from the letter of the word RECOVER?

Accepted Answer

Possible arrangements are 7!/2! 2! = 1260 [Division by 2 times 2! is because of the repletion of E and R .]

Question 13

There are 20 books of which 4 are single volumes and the other are books of 8, 5 and 3 volumes respectively. In how many ways can all these books be arranged on a self so that volumes of the same book are not separated?

Accepted Answer

Volumes of the same book are not to be separated i, e. all volumes of the same book are to be kept together, Regarding all volumes of the same book as one book, we have only 4 + 1 + 1 + 1 = 7 books. These seven books can be arranged in 7! ways. The book having 8 volumes can be arranged among themselves in 8! ways, the book having 5 volumes can be arranged among themselves in 5! ways, and the book having 3 volumes can be arranged among themselves in 3! ways. ∴ Required number = 7! 8! 5! 3!

Question 14

A library has two books each having three copies and three other books each having two copies. In how many ways can all these books be arranged in a shelf so that copies of the same book are not separated?

Accepted Answer

Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way. ∴ Required number = 5! x 1! x 1! x 1! x 1! = 120

Question 15

4 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated?

Accepted Answer

Assume the 2 girl students to be together i,e (one). Now there are 5 students. Possible ways of arranging them are 5! = 120 Now they (two girl) can arrange themselves in 2! ways. Hence, total ways = 120 x 2! = 240

Question 16

The total number of words, which can be formed out of the letters a, b, c, d, e, f taken 3 together, such that each word contains at least one vowel is?

Accepted Answer

The required number of words is : (2C1 x 4C2 + 2C2 x 2C1) 3! = 96

Question 17

A man has 5 friends and his wife has 4 friends. They want to invite either of their friends, one or more to a party. In how many ways can they do so?

Accepted Answer

Number of ways of selecting one or more friends from 5 friends = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 5 + 10 + 10 + 5 + 1 = 31 ways Number of ways of selecting one or more friends from 4 friends = 4C1 + 4C2 + 4C3 + 4C4 = 4 + 6 + 4 + 1 = 15 ways ∴ Total number of ways = 31 + 15 = 46 ways

Question 18

If there are 10 positive real numbers n 1 < n 2, n 3 .... < n 10 How many triplets of these numbers (n 1, n 2, n 3) (n 2, n 3, n 4), ... can be generated such that in each triplet the first number is always less than the second number and the second…

Accepted Answer

Three numbers can be selected and arranged out of 10 numbers in 10P3 ways 10!/7! = 10 x 9 x 8 Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways. Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2) = 120 ways

Question 19

How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

Accepted Answer

Total number of password using all alphabets -Total number of password using no symmetric alphabets = (26 x 25 x 24 ) - (15 x 14 x 13 ) = 12870

Question 20

A question paper had 10 questions. Each question could only be answered as True (T) or false (F). Each candidate answered all the questions. Yet, no two candidate wrote the answers in an identical sequence. How many different sequences of answers ar…

Accepted Answer

Each question can be answered in 2 ways. ∴ 10 question can be answered = 210 = 1024 ways

Permutation and Combination Questions