Permutation and Combination Questions for CAT with Answers PDF

The number of ways is the same as the number of ways in which m + n + p things can be divided into three groups containing m, n and p, which is 
∴ The required number of different allotments 

Different ways of arranging letters so that the positions of vowels and consonants remain unchanged 

Suppose there are 3 boxes B1, B2, B3
In total there are 8 boxes we are going to put 1 ball in each box
So, the remaining number of balls after putting 1 ball in each box = 8 - 3 = 5
so number of remaining balls n = 5
the number of boxes r = 3
The number of ways for arranging the balls in boxes without any condition = (n + r - 1)Cr-1
n + r - 1 = 5 + 3 - 1 = 7
r - 1 = 3 - 1 = 2

AIUA -- Vowels and 'PTLPTR' -- Consonants
Number of words that can be formed without changing the relative positions of vowels and consonants

= 2160

Out of the 36 possible outcomes, 12 outcomes will yield a score divisible by 3, i.e. a score of either 3, 6, 9 or 12.
The 12 outcomes will be various arrangements of (1, 2), (2, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (4, 5), (5, 4), (5, 1),(1, 5) and (6, 6).

Using the fundamental principle of multiplication, total number of types = 10 × 6 × 4 × 4 = 960

Since we are taking three letters at a time, the total number of three letter words is ⁶P₄ = 6 × 5 × 4 = 120. The number of words not containing A and E is ⁴P₃ = 4 × 3 × 2 = 24. Therefore, the required number of words is 96.

The first coin can be placed in any box out of 64 boxes.
Second coin cannot be in the same row or column. So, there should be 64 - (8 + 7) = 49 squares left, and so on.
Required number of ways = 64 × 49 × 36 × 25 × 16 = (6720)²

There are 6 men and 8 women.

Required number of ways (Number of men > Number of women) = C(6, 3) x C(8, 1) + C(6, 4)
So, the required probability 


Hence, option (1) is the correct answer.

There are 8 men and 6 women. Two men are fixed, so we have 6 men and 6 women for selection. Now, out of 9 persons, 2 men are fixed, so we have to select the rest 7 persons, including at least 1 man and 2 women.

Possible cases are as follows.
(1) 1 man, 6 women = C(6, 1) × C(6, 6)
(2) 2 men, 5 women = C(6, 2) × C(6, 5)
(3) 3 men, 4 women = C(6, 3) × C(6, 4)
(4) 4 men, 3 women = C(6, 4) × C(6, 3)
(5) 5 men, 2 women = C(6, 5) × C(6, 2)
So, total number of ways = [C(6, 1) × C(6, 6) + C(6, 2) × C(6, 5) + C(6, 3) × C(6, 4) + C(6, 4) × C(6, 3) + C(6, 5) × C(6, 2)]

= 6 + 90 + 300 + 300 + 90 = 786