# Permutation and Combination Questions for CAT Set-3 PDF

0
481

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.

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

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

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.

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.

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

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