Question

In how many ways can be letters of the word $PERMUTATIONS$ be arranged if the words start with $P$ and end with $S$.

Answer 1

In the word PERMUTATION, there are $12$ letters of which $T-2$

When P and S are fixed in first and last places, we have to arrange remaining letters in $10$ places can be done in $\frac{10!}{2!}$ ways

$\therefore$ Required number of words

$\frac{10!}{2!}$

$=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2!}{2!}$

$=10\times 9\times 8\times 7\times 6\times 5\times 4\times 3$

$=1814400$