Permutations and Combinations

1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 24400	B. 21300
C. 210	D. 25200

Explanation :

Number of ways of selecting 3 consonants out of 7 = ⁷C₃
Number of ways of selecting 2 vowels out of 4 = ⁴C₂

Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = ⁷C₃ x ⁴C₂

=(7×6×53×2×1)×(4×32×1)=210

It means that we can have 210 groups where each group contains total 5 letters(3 consonants

and 2 vowels).



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

= 5 x 4 x 3 x 2 x 1 = 120



Hence, Required number of ways = 210 x 120 = 25200

2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?


A. 159

B. 209


C. 201

D. 212

Explanation :

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.

Hence we have 4 choices as given below

We can select 4 boys ------(Option 1). 
Number of ways to this = 6C4

We can select 3 boys and 1 girl ------(Option 2)
Number of ways to this = 6C3 x 4C1

We can select 2 boys and 2 girls ------(Option 3)
Number of ways to this = 6C2 x 4C2

We can select 1 boy and 3 girls ------(Option 4)
Number of ways to this = 6C1 x 4C3

Total number of ways 
= (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3)
= (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) [Applied the formula nCr = nC(n - r) ] 

=[6×52×1]+[(6×5×43×2×1)×4]+[(6×52×1)(4×32×1)]+[6×4]
= 15 + 80 + 90 + 24

= 209



3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A. 624

B. 702


C. 756

D. 812




Explanation :

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.

Hence we have the following 3 choices

We can select 5 men ------(Option 1)
Number of ways to do this = 7C5

We can select 4 men and 1 woman ------(Option 2) 
Number of ways to do this = 7C4 x 6C1

We can select 3 men and 2 women ------(Option 3) 
Number of ways to do this = 7C3 x 6C2 

Total number of ways 
= 7C5 + [7C4 x 6C1] + [7C3 x 6C2]
= 7C2 + [7C3 x 6C1] + [7C3 x 6C2] [Applied the formula nCr = nC(n - r) ]

=[7×62×1]+[(7×6×53×2×1)×6]+[(7×6×53×2×1)×(6×52×1)]
= 21 + 210 + 525 = 756




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


A. 610

B. 720


C. 825

D. 920




Explanation :

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels 
should always come together. Hence these three vowels can be grouped and considered as a
single letter. That is, PTCL(OIA). 

Hence we can assume total letters as 5. and all these letters are different.
Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120

All The 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6

Hence, required number of ways = 120 x 6 = 720




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


A. 47200

B. 48000


C. 42000

D. 50400



Explanation :

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and 
these 5 vowels should always come together. Hence these 5 vowels can be grouped 
and considered as a single letter. that is, CRPRTN(OOAIO). 

Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and
rest of the letters are different.

Number of ways to arrange these letters = 7!2!=7×6×5×4×3×2×12×1=2520
In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.



Number of ways to arrange these vowels among themselves = 5!3!=5×4×3×2×13×2×1=20
Hence, required number of ways = 2520 x 20 = 50400