Permutation and Combination Questions for CAT 2026 — Derangement Problems with Solutions

PERMUTAION AND COMBINATION Question for CAT

Q.1 In how many ways, can 8 different balls be placed in 3 different boxes, when each bag can hold any number of balls?

a.8³

b.3⁸

c.8! / 3!

d.8! / 5!

Q.2 In how many ways can 5 different books be put inside in 4 different school bags?

a.5⁴

b.4⁵

^c.5P₄

d.None of the above

Q.3 Sum of three Natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

a.45

b.36

c.54

d.28

Q.4 Sum of three Whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

a.66

b.78

c.72

d.56

Q.5 n how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?

a.72

b.54

c.45

d.36

Q.6 Mohan, a thief, went to a liquor shop where he decided to take away 15 bottles. In the shop there are bottles of Wine, Whisky, Rum, Vodka and Gin. In how many ways he can select the bottles?

^a.19C₄

^b.20C₄

^c.15C₅

d.2200

Q.7 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

Q.8 Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. The number of ways in which we can place the balls in the boxes (order is not considered in the box) so that no box remains empty is.

a.150

b.41

c.50

d.None of the above

Q.9 Number of ways of distributing 9 identical balls into 3 identical boxes?

a.9

b.10

c.12

d.13

Q.10 Five racquets needs to be placed in three boxes. Each box can hold all the five racquets. In how many ways can the racquets be placed in the boxes so that no box can be empty if all racquets are different but all boxes are identical?

a.24

b.25

c.26

d.27

Q.11 In how many ways can 6 boys be allotted into 5 rooms such that no room is empty and all 6 boys are accommodated?

a.6 * 5! ways

b.7 * 5! ways

c.3 * 3! ways

d.15 * 5! Ways

Q.12 What is the number of terms in the expansion of (a + b + c)n, n ∈ N?

a.n+1

b.2n

c.(n+1)(n+2)

d.(n+1)(n+2)/2

Q.13 In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

a.6

^b.7C₂

^c.9C₂

d.None of the above

Q.14 Find the number of non negative integral solutions of a + b + c + d <=15

^a.14C₃

^b.18C₃

^c.19C₄

d.None of the above

Q.15 In how many ways can 12 balls be selected from an unlimited supply of identical red, blue and green balls?

^a.14C₂

^b.11C₂

^c.12P₃

d.None of the above

Q.16 There are 4 letters and 4 addressed envelopes. If each letter is randomly placed in an envelope, then in how many ways can wrong choices be made?

a.4³

b.4! – 1

c.16

d.4⁴ -1

Q.17 There are 4 letters and 4 addressed envelopes. If each letter is randomly placed in an envelope, then the number of ways in which only one letter goes in the wrong envelope is

a.D(4)

^b.4C₁+ ⁴C₂+ ⁴C₃ + ⁴C₄

^c.4C₁ x 3!

d.0

Q.18 There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelope, is

a.2

b.9

c.44

d.None of the above

Q.19 In how many ways can you put 7 letters into their respective envelopes such that exactly 3 go into the right envelope?

a.315

^b.7C₃

c.D(4)

d.None of the above

Q.20 A postman can put 12 letters into their respective envelopes such that exactly 5 will go into the right envelope. Find the number of ways of doing this work

^a.12C₅D(7)

^b.12P₅.D(7)

^c.12C₅D(5)

^d.12P₇.D(5)

Answer for Permutation and Combination for CAT

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Frequently Asked Questions — Derangements for CAT 2026

What is a derangement?

A derangement is a permutation where no element appears in its original position. For example, for {1,2,3}: derangements are {2,3,1} and {3,1,2} — neither 1, 2, nor 3 is in its original position. Denoted D(n) or !n.

What is the formula for the number of derangements of n objects?

D(n) = n! x (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!). For small n: D(1)=0, D(2)=1, D(3)=2, D(4)=9, D(5)=44. A useful approximation: D(n) ≈ n!/e for large n, where e ≈ 2.718.

What is the recurrence relation for derangements?

D(n) = (n-1) x [D(n-1) + D(n-2)]. This comes from considering whether element n swaps with one other element or participates in a longer cycle. It allows easy calculation without the full inclusion-exclusion formula.

How are derangements tested in CAT?

CAT tests derangements in questions like: 'In how many ways can n letters be placed in n envelopes such that no letter goes into the correct envelope?' The answer is D(n). It may also appear in probability questions.

What is the probability that a random permutation is a derangement?

P(derangement) = D(n)/n! ≈ 1/e ≈ 0.368 for large n. This is a useful result that CAT sometimes tests in probability form: 'What is the probability that none of the n elements is in its original position?'

How do I solve partial derangement problems?

Partial derangements fix some elements in their correct positions and derange the rest. If exactly k elements are fixed, the number of ways = C(n,k) x D(n-k). This extends the basic derangement formula.

What is the inclusion-exclusion principle as applied to derangements?

D(n) = n! - C(n,1)(n-1)! + C(n,2)(n-2)! - ... = sum over k from 0 to n of (-1)^k x n!/k!. This counts total permutations minus those with at least one element in its original position.

How should I prepare derangements for CAT?

Memorise D(n) values for n=1 to 5 and the recurrence relation. Practice 5-8 problems covering letter-envelope type, probability of derangement, and partial derangements. This topic appears in 1 CAT question every 2-3 years.