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:

In how many ways can you choose one or more of 12 different candies?

A) 4054 B) 4050
C) 4095 D) 4059
 
Answer & Explanation Answer: C) 4095

Explanation:

Each candy can be dealt with in two ways.It can be chosen or not chosen.This will give 2 possibilites for the first candy, 2 for the second, and so on.by multiplying the cases together we get 212. Since the case of no candy being selected is not an option, we have to subtract 1 from our answer.

 

Therefore,there are 212- 1 = 4095 ways of selecting one or more candies

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0 8385
Q:

In how many different ways can 6 different balls be distributed to 4 different boxes, when each box can hold any number of ball?

A) 2048 B) 1296
C) 4096 D) 576
 
Answer & Explanation Answer: C) 4096

Explanation:

Every ball can be distributed in 4 ways. 

Hence the required number of ways = 4 x 4 x 4 x 4 x 4 x 4 = 4096

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0 8240
Q:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?

A) 15 B) 20
C) 5 D) 10
 
Answer & Explanation Answer: B) 20

Explanation:

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.

 

So, there is one way of doing it.

 

Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.

 

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

 

Required number of numbers = (1 x 5 x 4) = 20.

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12 8171
Q:

What is the value of 100P2 ?

A) 10000 B) 9900
C) 8900 D) 7900
 
Answer & Explanation Answer: B) 9900

Explanation:

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

 

That can be done as,

 

100 P2  = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

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8 8153
Q:

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is  ?

A) 2(6!) B) 6! x 7
C) 6! x ⁷P₆ D) None
 
Answer & Explanation Answer: C) 6! x ⁷P₆

Explanation:

We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.

Hence required number of ways = 6! x ⁷P₆

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6 7999
Q:

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

A) 360 B) 420
C) 576 D) 220
 
Answer & Explanation Answer: A) 360

Explanation:

No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360

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4 7969
Q:

From a bunch of flowers having 16 red roses and 14 white roses, four flowers have to be selected. In how many different ways can they be selected such that at least one red rose is selected?

A) 27405 B) 26584
C) 26585 D) 27404
 
Answer & Explanation Answer: D) 27404

Explanation:

Given total 16 Red roses and 14 White roses = 30 roses

 

Four flowers have to be selected from 30 i.e,  C430= 27405 Ways 

 

Now, atleast one Red rose is selected i.e, 27405(total) - 1(all four are white roses)  = 27404 ways. 

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10 7961
Q:

How many four digit even numbers can be formed using the digits {2, 7, 5, 3, 9, 1} ?

A) 59 B) 60
C) 61 D) 64
 
Answer & Explanation Answer: B) 60

Explanation:

The given digits are 1, 2, 3, 5, 7, 9  

A number is even when its units digit is even. Of the given digits, two is the only even digit.Units place is filled with only '2' and the remaining three places can be filled in ⁵P₃ ways.

 

Number of even numbers = ⁵P₃ = 60.

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1 7795