Question

Find the number of words which can be formed out of the letters $a,b,c,d,e,f$ taken $3$ together, each containing one vowels at least.

Answer 1

a,e - vowels - 2 nos

b,c,d,f - constant - 4 nos

Case 1 : 1 vowel is selected

out of 2 vowels, 1 can be selected in

${}^2{C}_1$ ways =2

out of 4 constants, 2 constants can be selected in ${}^4{C}_2=\frac{4\times 3}{2\times 1}=6$ ways.

All three letters can arrange among themselves in 3! ways

Thus total possibility = $2\times 6\times 3!$

$=2\times 6\times 6=72$

Case 2 : 2 vowels is selected

2 vowels can be selected in ${}^2{C}_2$ ways = 1

1 constant can be arranged in 3! ways

Thus,

total possibility = $1\times 4\times 3!$

=$1\times 4\times 6=24$

Thus total number of arrangements

= 72+24

=96