Question

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

• A. $1814400$
• B. $1814405$
• C. $1824050$
• D. $Noneofthese$

Answer 1

The correct option is A $1814400$

Let first position be $P$ and last position be $S$(both are fixed)

Since letter are repeating

Hence we use this formula $\frac{n!}{p_1!p_2!p_3!}$

Total number of letters$=n=10$

and since, $2T$

$\Rightarrow p_1=2$

Total arrangements$=\frac{10!}{2!}=181440$