Mahindra Comviva Permutation and Combinations Problems with Solutions Quiz -1

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
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!=120
Hence, required number of ways
210*120=25200

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 belowWe 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

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 choicesWe 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

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] = 120All 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

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

We need to select 5 men from 7 men and 2 women from 3 womenNumber of ways to do this
= 7C5 x 3C2
= 7C2 x 3C1 [Applied the formula nCr = nC(n – r) ]
=(7×62×1)×3
= 21 x 3 = 63

The word ‘MATHEMATICS’ has 11 letters. It has the vowels ‘A’,’E’,’A’,’I’ in it and
these 4 vowels must always come together. Hence these 4 vowels can be grouped and
considered as a single letter. That is, MTHMTCS(AEAI).Hence we can assume total letters as 8. But in these 8 letters, ‘M’ occurs 2 times,
‘T’ occurs 2 times but rest of the letters are different.
Hence,number of ways to arrange these letters = 8!(2!)(2!)=8×7×6×5×4×3×2×1(2×1)(2×1)=10080
In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.



Number of ways to arrange these vowels among themselves = 4!2!=4×3×2×12×1=12
Hence, required number of ways = 10080 x 12 = 120960

We need to select 5 men from 8 men and 6 women from 10 womenNumber of ways to do this
= 8C5 x 10C6
= 8C3 x 10C4 [Applied the formula nCr = nC(n – r) ]
=(8×7×63×2×1)(10×9×8×74×3×2×1)
= 56 x 210

= 11760

The word ‘LOGARITHMS’ has 10 different letters.Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= 10P3
= 10 x 9 x 8
= 720

The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and
these 3 vowels should always come together. Hence these 3 vowels can be grouped
and considered as a single letter. that is, LDNG(EAI).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 = 120In the 3 vowels (EAI), all the vowels 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