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:

Prn=nn-1n-2....n-r+1=n!n-r!

 

Ex : (i) P26=6×5=30   (ii) P37=7×6×5=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 p1+p2+...+pr=n

Then, number of permutations of these n objects is :

n!(p1!)×(p2! ).... (pr!)

 

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:

Crn=n!(r !)(n-r)!=nn-1n-2....to r factorsr!

 

Note : (i)Cnn=1 and C0n =1     (ii)Crn=C(n-r)n

 

Examples : (i) C411=11×10×9×84×3×2×1=330      (ii)C1316=C(16-13)16=C316=560

Q:

If each vowel of the word NICELY is changed to the next letter in the English alphabetical series and each consonant is changed to the previous letter in the English alphabetical series and then the alphabets are arranged in alphabetical order, which of the following will be fourth from the left?

A) M B) J
C) B D) O
 
Answer & Explanation Answer:

Explanation:

Thus after arranging the letters as per English alphabetical series; we get; Thus 4th letter from the left end will be K.

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16 2160
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
 
Answer & Explanation Answer: 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

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7 2063
Q:

In how many different ways could couples be picked from 6 men and 9 women ?

A) 26 B) 54
C) 52 D) 28
 
Answer & Explanation Answer: B) 54

Explanation:

Number of mens = 8

Number of womens = 5

 

Different ways could couples be picked = C16×9C1 = 9 x 6 = 54 ways.

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5 2049
Q:

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

A) 720 B) 5040
C) 360 D) 180
 
Answer & Explanation Answer: A) 720

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

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11 1851
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
 
Answer & Explanation Answer: A) 120

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.

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8 1793
Q:

How many such pair of letters are there in the word ‘TROUBLED’ which have as many letters between them in the word as they have between them in the English alphabet?

A) 2 B) 3
C) 4 D) 5
 
Answer & Explanation Answer: A) 2

Explanation:

Hence there are two pairs

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13 1741
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
 
Answer & Explanation Answer: B) 1440

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.

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13 1725
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
 
Answer & Explanation Answer: 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 = 

2652 + 452 = 3052 = 1526

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10 1693