Question

Consider all permutations of the letters of the word MORADABAD.

The number of permutations which contain the word BAD is:

• A. $21\times 5!$
• B. $7\times 5!$
• C. $6\times 5!$
• D. $2\times 5!$

Answer 1

The correct option is A $21\times 5!$

Total number of letters = $9$

Consider BAD as a group, then the number of total letters = $6+1=7$

$\therefore$ Permutation of 7 things = $7!$

but letters remaining after removing BAD will have two A's

$\therefore$ Total permutation = $\frac{7!}{2}$ = $21\cdot 5!$