Permutation and Combination for XAT

Permutation and Combination for XAT
Permutation and Combination for XAT

Permutation and Combination for XAT

Download important Permutations and Combination Questions for XAT PDF based on previously asked questions in the XAT exam. Practice Permutations and Combination Questions PDF for the XAT exam.

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Question 1: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and 0 so as to form a triangle?

Question 2: In how many ways is it possible to choose a white square and a black square on a chessboard so that the squares must not lie in the same row or column?

a) 56

b) 896

c) 60

d) 768

Question 3: How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

a) 499

b) 500

c) 375

d) 376

Question 4: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

a) 16

b) 20

c) 14

d) 15

Question 5: The number of ways in which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same game is:

a) 1514

b) 1512

c) 3024

d) None of the above

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Question 6: In how many ways can eight directors, the vice chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the the vice chairman and a specific director?

a) 9! × 2

b) 2 × 8!

c) 2 × 7!

d) None of these

Question 7: Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?

a) 700

b) 504000

c) 3528000

d) 7056000

Question 8: A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there has to be exactly 3 girls in the invitees?

a) 320

b) 160

c) 80

d) 200


Directions for the next two questions: Answer the questions based on the following information.

Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters.

Question 9: How many four-letter computer passwords can be formed using only the symmetric letters (no repetition allowed)?

a) 7,920

b) 330

c) 14,640

d) 4,19,430

Question 10: How many three-letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?

a) 990

b) 2,730

c) 12,870

d) 15,600

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Answers & Solutions:

1) Answer: 160

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 – 5 = 160

2) Answer (D)

First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosen in 24 ways. So total no. of ways of selection is 32*24 = 768.

3) Answer (D)

We have to essentially look at numbers between 1000 and 4000 (including both).

The first digit can be either 1 or 2 or 3.

The second digit can be any of the five numbers.

The third digit can be any of the five numbers.

The fourth digit can also be any of the five numbers.

So, total is 3*5*5*5 = 375.

However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376

4) Answer (A)

We have been given that a + b + c + d = 7
Total ways of distributing 7 things among 4 people so that each one gets at least one = $^{n-1}C_{r-1}$ = 6C3 = 20
Now we need to subtract the cases where any one person got more than 3 erasers. Any person cannot get more than 4 erasers since each child has to get at least 1. Any of the 4 childs can get 4 erasers. Thus, there are 4 cases. On subtracting these cases from the total cases we get the required answer. Hence, the required value is 20 – 4 = 16

5) Answer (B)

There are 9 married couples. Therefore, there will be 9 men and 9 women.
First, let us select the 2 men.
2 men can be selected in 9C2 = 36 ways.

Now, the wives of these 2 men cannot be selected. Therefore, we have to select 2 women from the remaining 7 women.
2 women can be selected in 7C2 = 21 ways.

We have selected 2 men and 2 women. A team should consist of one man and one woman. Therefore, the 2 teams can be formed in 2 ways.

Therefore, the total number of ways in which the team can be selected is 36*21*2 = 1512.
Therefore, option B is the right answer.

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6) Answer (C)

Chariman, Vice-Chairman and the director can be made as a group such that Chairman sits between the Vice-Chairman and the director. This group can be formed in 2 ways.

Each of the remaining 7 directors and the group can be arranged in 7! ways.

=> Total number of ways = 2 * 7!.

7) Answer (C)

Out of 8 consonants and 5 vowels selecting 4 consonants and 3 vowels can be done in $^8C_4 * ^5C_3 = \dfrac{8*7*6*5*5*4}{4*3*2*2}$
= 700
Internal arrangement of 7 letters = 7!
Thus, the total number of ways = 7!*700 = 352800
Hence, option C is the correct answer.

8) Answer (B)

Selecting 3 girls from 5 girls can be done in $^5C_3$ ways => 10 ways

Each of the boys may or may not be selected => 2 * 2 * 2 * 2 = 16 ways

=> 16 * 10 = 160 ways

9) Answer (A)

The number of ways in which this can be done is 11*10*9*8 = 7920

10) Answer (C)

If there are 3 symmetric letters, it can be formed in 11*10*9 ways

If there are 2 symmetric letters, it can be formed in 11C2 * 15C1 * 3! ways

If there is only 1 symmetric letter, the password can be formed in 15C2*11C1*3! ways

Total = 990+330*15+630*11 = 12870 ways

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We hope this Permutation and Combination  Questions PDF for XAT with Solutions will be helpful to you.


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