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:

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

A) 600 B) 610
C) 609 D) 599
 
Answer & Explanation Answer: A) 600

Explanation:

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways = 600 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

6 13617
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

Report Error

View Answer Report Error Discuss

9 13495
Q:

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together?

A) 1440 B) 720
C) 2250 D) 3600
 
Answer & Explanation Answer: D) 3600

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

These 7 letters can be arranged in 7! ways.

Number of vowels in the given word = 2 (E, A)

The number of ways of arrangement in which vowels come together is 6! x 2! ways

 

Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.

 

Report Error

View Answer Report Error Discuss

11 13449
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

Report Error

View Answer Report Error Discuss

6 13201
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

Report Error

View Answer Report Error Discuss

19 13163
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

Report Error

View Answer Report Error Discuss

4 13147
Q:

If there are 15 dots on a circle,how many triangles can be formed?

A) 455 B) 450
C) 469 D) 500
 
Answer & Explanation Answer: A) 455

Explanation:

There are 15 dots in total,and to make a triangle we need to select any three of those dots.

 

So,15C3 = 455

Report Error

View Answer Report Error Discuss

5 12644
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

Report Error

View Answer Report Error Discuss

9 12532