# 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:

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

 A) 17756 B) 17576 C) 12657 D) 12666

Explanation:

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways

The first two letters can chosen in 26 x 26 = 676 ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

All the three letters can be chosen in 676 x 26 =17576 ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

8 10918
Q:

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

 A) 500 B) 490 C) 495 D) 540

Explanation:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is $12C4$ = 495.

Therefore, we can draw 495 quadrilaterals

0 10793
Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

 A) 209 B) 290 C) 200 D) 208

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = $6C1*4C3+6C2*4C2+6C3*4C1+6C4$

$6C1*4C1+6C2*4C2+6C3*4C1+6C2$ = 209.

3 10737
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

23 10700
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

9 10551
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.

8 10467
Q:

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

 A) 2400 B) 6400 C) 3600 D) 7200

Explanation:

Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1.

Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

Case 2 : Anju is allotted bed number 7.

Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

10 10465
Q:

The number of ways that 8 beads of different colours be strung as a necklace is

 A) 2520 B) 2880 C) 4320 D) 5040

The number of ways of arranging n beads in a necklace is $(n-1)!2=(8-1)!2=7!2$ = 2520