Question

If the total number of arrangements of letters $a^2{b}^3{c}^4$ when written at full length is $N$, then sum of digits of $N$ is

Answer 1

Given letter : $a^2{b}^3{c}^4$,
Wen written at full length $\Rightarrow aabbbcccc$
Total letters $=9(2a^{\prime }s+3b^{\prime }s+4c^{\prime }s)$
$\therefore$ Total arrangements
$N=\frac{9!}{2!\cdot 3!\cdot 4!}=1260$
$\therefore$ Sum of digits of $N$ is $9$