# Permutations and Combinations Questions

FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

1.  Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.

Examples : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

2.  Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$

Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$   (ii) $P_{3}^{7}=\left(7×6×5\right)=210$

Cor. number of all permutations of n things, taken all at a time = n!.

Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$

Then, number of permutations of these n objects is :

3.  Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Note : (i)     (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$

Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$      (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$

Q:

How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?

 A) 120 B) 240 C) 256 D) 360

Explanation:

Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are

5 x 4 x 3 x 2 x 1 = 120.

0 14
Q:

What is the value of ?

 A) 10000 B) 9900 C) 8900 D) 7900

Explanation:

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

= 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99

= 9900.

1 492
Q:

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?

 A) 720 B) 1440 C) 1800 D) 3600

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

Number of vowels in the given word = 2 = A & E

Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is

6! x 2! = 720 x 2 = 1440.

6 363
Q:

In how many different ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together?

 A) 112420 B) 85120 C) 40320 D) 1209600

Explanation:

Given word is TRANSFORMER.

Number of letters in the given word = 11 (3 R's)

Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is

10! x 2!/3!

= 3628800 x 2/6

= 1209600

2 369
Q:

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

 A) 1,51,200 ways. B) 5,04,020 ways C) 72,000 ways D) None of the above

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways

Remaining 7 letters can be arranged in 7!/3! x 2! ways

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

2 475
Q:

In how many different ways can the letters of the word 'RITUAL' be arranged?

 A) 720 B) 5040 C) 360 D) 180

Explanation:

The number of letters in the given word RITUAL = 6

Then,

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

=> 6 x 5 x 4 x 3 x 2 x 1 = 720

3 436
Q:

How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.

 A) 64 B) 63 C) 62 D) 60

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

6 702
Q:

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king?

 A) 15/52 B) 17/26 C) 13/17 D) 15/26

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

Required, the probability that if a card is drawn either card is black or a king =