Question

Find the number of ways in which the letter of the word ${}^{\prime }MADHUBAN{I}^{\prime }$ be arranged if
$\left(i\right)$ vowels occupy odd places
$\left(ii\right)$ all the vowels are never together
$\left(iii\right)$ words do not begin with $M$ but end with $I$.

Answer 1

(i) vowels occupy odd places

There are only 2 vowels and 5 odd places are there.

Therefore the vowels occupying odd places $=5\times 2=10$

(ii) all the vowels are never together

Consider A in 1st place; and last place I.
Balance 7 places, 7 letters (M, B, D, H, U, N) =$1\times 2\times 7!=5040$

(iii) words do not begin withMbut end withI

Consider any one from the 5 letters - B, D, H, U, N in the 1st place
and I in the last place:

So first place = 5 ways and last place 1 way $=5\times 1=5$
Rest 7 places, two As and any one from the 5 letters $=\frac{7!}{2!}=2520$
So its arrangement $=21520\times 5=12600$

Thus total permutation $=5040+12600=17640$