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:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 A) 25200 B) 52000 C) 120 D) 24400

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ($7C3$*$4C2$

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

283 152873
Q:

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

 A) 720 B) 520 C) 700 D) 750

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

89 87349
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

 A) 196 B) 186 C) 190 D) 200

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in  $4C3*6C2$  ways

or, 4 women, 1 man : It can be done in  $4C4*6C1$ways

or, 2 women, 3 men : It can be done in $4C2*6C3$ ways.

Total number of ways of forming the committees

$4C2*6C3+4C3*6C2+4C4*6C1$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

94 67958
Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

 A) 1260 B) 1400 C) 1250 D) 1600

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in $10C6$ or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

57 58875
Q:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

 A) 135 B) 63 C) 125 D) 64

Explanation:

Required number of ways =

55 58011
Q:

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

 A) 216 B) 45360 C) 1260 D) 43200

Explanation:

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = $9!2!×2!×2!$ = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $6!2!×2!$ = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $4!2!$ = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

64 57457
Q:

12 people at a party shake hands once with everyone else in the room.How many handshakes took place?

 A) 72 B) 66 C) 76 D) 64

Explanation:

There are 12 people, so this is our n value.

So, $12C2$1= 66

100 50829
Q:

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

 A) 360 B) 700 C) 720 D) 120

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.