Question

In how many ways can the letters of the word $PERMUTATIONS$ be arranged if there are always $4$ letters between $P$ and $S$?

Answer 1

Positions P and S are fix.

We need to arrange remaining letters,

(E,R,M,U,T,A,T,I,O,N)

Since, T is repeated,

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

Number of letters $=10$

$n=10$

Here 2T's

$p_1=2$

No of arrangements $=\frac{10!}{2!}$

Hence,

Number of arrangements when P is before S $=7\times \frac{10!}{2!}$

Similarly,

S can be before P

Number of arrangements when S is before P $=7\times \frac{10!}{2!}$

Total number of arrangements $=7\times \frac{10!}{2!}+7\times \frac{10!}{2!}$

$=2\times 7\times \frac{10!}{2!}$

$=7\times 10!$

$=2540100$.