**Permutation and Combination Questions For CAT Set-3 PDF:**

Download Permutation and Combination Questions For CAT Set-3 PDF. Practice important problems with detailed explanations for Permutation and Combination Questions related to conversions and sums.

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**Question 1: **The number of ways of reaching origin from point (3,3) such that a person can take only one step at a time either along a x axis or along a y axis is

a) 28

b) 20

c) 45

d) 55

**Question 2: **There are five prize winners – A, B, C, D and E – and 7 prizes – First, second, third, fourth, fifth, sixth and seventh. In how many ways can these prizes be distributed such that all the prizes that A get are better than those that B gets, all the prizes that B get are better than those that C gets and so on (Everyone gets at least 1 prize?

a) 10

b) 15

c) 17

d) 18

**Question 3:** Five different prizes are to be distributed among 3 students A,B,C such that each student gets at least 1 prize. Find the total number of ways in which prizes can be distributed?

a) 130

b) 100

c) 150

d) 120

**Question 4: **In how many ways can 18 identical plates be divided into 3 groups such that each group must at least have one plate?

a) 20

b) 24

c) 25

d) 27

**Question 5:** In how many ways can you distribute 20 identical chocolates to 4 children such that each of them gets at least one?

a) 969

b) 900

c) 950

d) 960

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__Answers & Solutions:__

**1) Answer (A)**

There are 6 steps in total out of whih 3 are horizontal and 3 are vertical:HHHVVV.

Number of ways of reaching origin is = 6!/(3!*3!) = 20

**2) Answer (B)**

1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7

We need to put 4 persons in these 6 places such that all the prizes that come between a person X and the person before him belong to person X.

There are 6 gaps and 4 persons to be put. The remaining prizes to the right of the fourth person belong to the fifth person.

This can be done in $^6C_4$ = 15 ways.

**3) Answer (C)**

Since the prizes are different, the required distribution would take place in (1,1,3), (2,2,1) ways.

Let us consider the first arrangement – (1,1,3)

There are 3 ways in which the person who gets 3 prizes can be selected. The prizes to be distributed to him can be selected in 5C3 ways. Now the remaining 2 prizes can be distributed to the 2 persons in 2! ways.

Number of ways = 3*5C3*2 = 3 *10* 2 = 60 ways

Let us consider the second arrangement – (2,2,1)

There are 3 ways in which we can select the person who gets 1 prize. The prize for him can be selected in 5C1 = 5 ways. The remaining 2 persons get 2 prizes each. So, we have to select only the ways in which 2 prizes can be selected. 2 prizes can be selected from the 4 available prizes in 4C2 = 6 ways.

Number of ways = 3*5C1*6 = 3*5*6 = 90 ways.

Total number of ways = 60 + 90 = 150.

**4) Answer (D)**

Let the number of plates in the three groups be x, y and z.

If x = 1, then the remaining 17 plates can be distributed between y and z is 8 ways => (1,16), (2,15), (3,14), (4,13), (5,12), (6,11), (7,10), (8,9)

If x = 2, then the remaining 16 plates can be distributed between y and z is 7 ways => (2,14), (3,13), (4,12), (5,11), (6,10), (7,9), (8,8)

If x = 3, then the remaining 15 plates can be distributed between y and z is 5 ways => (3,12), (4,11), (5,10), (6,9), (7,8)

If x = 4, then the remaining 17 plates can be distributed between y and z is 4 ways => (4,10), (5,9), (6,8), (7,7)

If x = 5, then the remaining 17 plates can be distributed between y and z is 2 ways => (5,8), (6,7)

If x = 6, then the remaining 17 plates can be distributed between y and z is 1 ways => (6,6)

=> Total number of ways = 8 + 7 + 5 + 4 + 2 + 1 = 27

**5) Answer (A)**

Let the number of chocolates distributed to 4 children be a, b, c and d respectively.

a + b + c + d = 20

Each of them gets at least one => a + b + c + d = 16

Number of non-negative solutions to this equation is $^{16+4-1}C_{4-1}$ = $^{19}C_3$ = 969