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:

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

A) 120 B) 256
C) 192 D) 244
 
Answer & Explanation Answer: C) 192

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

 

The number can be 4 digited but greater than 4000 or 5 digited.

 

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x P34 = 3 x 24 = 72. 

 

5 digited numbers = P55 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

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21 14706
Q:

If repetition of the digits is allowed, then the number of even natural numbers having three digits is :

A) 550 B) 450
C) 500 D) 540
 
Answer & Explanation Answer: B) 450

Explanation:

In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450

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10 14684
Q:

Some children goto ice-cream shop. 9 flavours are available there. Each child takes a cone with two different flavours. No two children take same combination and they have taken all such possible combinations. How many children went to ice cream shop ?

A) 28 B) 56
C) 44 D) 36
 
Answer & Explanation Answer: D) 36

Explanation:

Given there are 9 flavours of ice creams.

 

Each child takes the combination of two flavours and no two children will have the same combination

 

This can be done by  ways i.e children.

 

Number of children=C29  = 9 x 8 / 1x 2 = 36.

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4 13949
Q:

In how many ways can 5 letters be posted in 4 letter boxes?

A) 512 B) 1024
C) 625 D) 20
 
Answer & Explanation Answer: B) 1024

Explanation:

First letter can be posted in 4 letter boxes in 4 ways. Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.

Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

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9 13837
Q:

If nC10=nC12  then,find n.

A) 10 B) 12
C) 22 D) 24
 
Answer & Explanation Answer: C) 22

Explanation:

Using, Crn=Cn-rn we get 

n – 10 = 12

or, n = 12 + 10 = 22

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19 13787
Q:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555 B) 106665
C) 106656 D) 108333
 
Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

 

 

 

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

 

 

 

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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9 13655
Q:

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are never together.

A) 120 B) 640
C) 720 D) 540
 
Answer & Explanation Answer: C) 720

Explanation:

There are 7 letters in the word ‘Bengali of these 3 are vowels and 4 consonants.

 

Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

 

Total number of words = 5! x 3!= 120 x 6 = 720

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6 13560
Q:

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

A) 48 B) 64
C) 63 D) 45
 
Answer & Explanation Answer: B) 64

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

 

Required number of ways=3C1*6C2+3C2*6C1+3C3 = (45 + 18 + 1) =64

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5 13453