Question

Find the number of arrangements that can be made out of the letters of the words
(1) independence, (2) superstitious, (3) institutions

Answer 1

$\left(1\right)$ Independence, $12$ letters, $4e^{\prime }s,3n^{\prime }s,2d^{\prime }s$
$\therefore$ Total words formed $=\frac{12!}{4!3!2!}=1663200$
$\left(2\right)$ Superstitious, $13$ letters, $3s^{\prime }s,2u^{\prime }s,2t^{\prime }s,2i^{\prime }s.$
Total no. of words formed $=\frac{13!}{3!\times 2!\times 2!\times 2!}=129729600$
$\left(3\right)$ Institutions, $12$ letters, $3i^{\prime }s,2n^{\prime }s,2s^{\prime }s,3t^{\prime }s.$
Total no. of words formed $=\frac{12!}{3!2!2!3!}=3326400$