Permutation and combination problems with solutions and answers PDF

Download Permutation and Combination Problems with solutions pdf. Today, I am going to share techniques to solve permutation and combination questions. This chapter talk about selection and arrangement of things which could be any numbers, persons,letters,alphabets,colors etc.The basic difference between permutation and combination is of order.

Right now there is fierce competition among candidates, it’s hard to crack any exam without scoring really good in Quantitative Aptitude.

 While analysing the papers from last many years, I noticed that less than a dozen of questions in every chapter are repeated in every exam.

Download Permutation and Combination Problems with solutions pdf (link given below post)

1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Answer With Explanation : 

Answer: Option D

Explanation:

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

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×53×2×1)×(4×32×1)=210=(7×6×53×2×1)×(4×32×1)=210

It means 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!=5×4×3×2×1=120=5!=5×4×3×2×1=120

Hence, required number of ways
=210×120=25200=210×120=25200

Answer With Explanation : 

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 options 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 × 4C1

We can select 2 boys and 2 girls …(option 3)
Number of ways to this = 6C2 × 4C2

We can select 1 boy and 3 girls …(option 4)
Number of ways to this = 6C1 × 4C3

Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]

=6×52×1+6×5×43×2×1×4=6×52×1+6×5×43×2×1×4 +6×52×1×4×32×1+6×4+6×52×1×4×32×1+6×4

=15+80+90+24=209=15+80+90+24=209

Answer With Explanation

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 options.

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 × 6C1

We can select 3 men and 2 women …(option 3)
Number of ways to do this = 7C3 × 6C2

Total number of ways
= 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 7C2 + (7C3 × 6C1) + (7C3 × 6C2)[∵ nCr = nC(n – r) ]

=7×62×1+7×6×53×2×1×6=7×62×1+7×6×53×2×1×6 +7×6×53×2×1×6×52×1+7×6×53×2×1×6×52×1

=21+210+525=756=21+210+525=756

Permutation and Combination Problems with solutions pdf (link given below )

Answer With Explanation

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 vowels among themselves =5!3!=5×4×3×2×13×2×1=20=5!3!=5×4×3×2×13×2×1=20

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