JEE Main · 2008 · 27 Apr 2008 · Offline · Permutations and Combinations

The word has eleven letters, with four I's, four S's, two P's and one M. First set down the seven non-S letters, namely $M$, the four I's and the two P's, which arrange in $\frac{7!}{4!2!}=105$ ways, and $105=7\cdot \left(\genfrac{}{}{0ex}{}{6}{4}\right)$. These seven letters leave eight gaps, and to keep the S's apart four of the gaps are chosen for them in $\left(\genfrac{}{}{0ex}{}{8}{4}\right)$ ways. The total count is therefore $7\cdot \left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{8}{4}\right)$.