Permutation and Combination Exercise

PERMUTATION AND COMBINATION EXERCISE :

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

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

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

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!=5×4×3×2×1=120=5!=5×4×3×2×1=120

All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6

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

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

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=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=5!3!=5×4×3×2×13×2×1=20

Hence, required number of ways
=2520×20=50400=2520×20=50400

4.  In how many ways can 5 man draw water from 5 taps if no tap can be used more than once?

Explanation

1^st man can draw water from any of the 5 taps.
2^nd man can draw water from any of the remaining 4 taps.
3^rd man can draw water from any of the remaining 3 taps.
4^th man can draw water from any of the remaining 2 taps.
5^th man can draw water from remaining 1 tap.

5

4

3

2

1

Hence total number of ways
=5×4×3×2×1=120