# 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:

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

 A) 564 B) 735 C) 756 D) 657

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways= $7C3*6C2+7C4*6C1+7C5$ = 756.

12 19070
Q:

Four ladies A, B, C and D and four gentlemen E, F, G and H are sitting in a circle round a table facing each other.

Directions:

(1) No two ladies or two gentlemen are sitting side by side.

(2) C, who is sitting between G and E is facing D.

(3) F is between D and A and is facing G.

(4) H is to the right of B.

Question:

1. Who are immediate neighbours of B?

2. E is facing whom?

 A) G & H , H B) F & H , B C) E & F , F D) E & H , G

Explanation:

From the directions given :

From the fig. it is clear that

1) neighbours of B are G , H.

2) E is facing H.

24 17814
Q:

How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?

 A) 215 B) 315 C) 415 D) 115

Explanation:
Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.

Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.

Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315

30 17589
Q:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

 A) 126 B) 120 C) 146 D) 156

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore, the total number of ways in which 8 students can travel is:
$8C3+8C4$=56 + 70= 126

13 17496
Q:

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

 A) 1170 B) 1200 C) 720 D) 360

Explanation:

Possibilities     Bowlers      Batsmen         Number of ways

6               9

1              4                7              $(6C4*9C7)$

2              5                6              $6C5*9C6$

3              6                5              $6C6*9C5$

$6C4*9C7$ = 15 x 36 = 540

$6C5*9C6$ = 6 x 84 = 504

$6C6*9C5$= 1 x 126 = 126

Total = 1170

15 17027
Q:

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

 A) 180 B) 220 C) 240 D) 160

Explanation:

Number of ways of choosing 2 black pens from 5 black pens in $5C2$ ways.

Number of ways of choosing 2 white pens from 3 white pens in $3C2$ ways.

Number of ways of choosing 2 red pens from 4 red pens in $4C2$ways.

By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

12 16372
Q:

 A) 18! B) 18! x 19! C) 18!(6 x 24) D) 18! x 30

Explanation:

Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular–permutations: $18P122*12$ = $18!6*24$

35 16313
Q:

Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

 A) 340 B) 370 C) 320 D) 330

Explanation:

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in 11$C4$ ways = 330 ways