Aptitude Questions and Answers

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

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Answer : Option D

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

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Answer : Option B

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 ⁿC_r = 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

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Answer : Option C

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 ⁿC_r = 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

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Answer : Option B

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

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Answer : Option D

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