In how many ways can the letters of word $PERMUTATIONS$ be arranged if there are always $4$ letters between $P$ and $S$?
Now there are $12$ blank spaces, which we have to fill by the letters of the given word.
That is :-
[][][][][][][][][][][][]
it is given that two places are always filled with P and S such that there are 4 blank spaces between them.
{[P][][][][][S]}[][][][][][]
$Therefore$, the number of arrangements that can be made such that there are 4 letter between $P$ and $S$ is $P(10,4)$
$p(10, 4) = \frac{10!}{(10 – 4)!} =\frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040 \tag 1$
Also {[P][][][][][S]} can be arranged in different ways with respect to {[P][][][][][S]}[][][][][][].
$Therefore$ the number arrangements will be $p(7, 7)$
$p(7, 7,) = 7! = 5040 \tag 2$
$Therefore$ the the total number of arrangements for the given question will be
$5040 \times 5040 \text{ (results of 1 and 2)}\\
= 25401600 \tag3$
but there are two T's in the word permutation
$Therefore$ we divde the result of 3 by 2 to get the real answer.
That is , $$\frac{25401600}{2} = 12700800$$
So I am getting $12700800$ as the answer but it is wrong the given answer is $25401600$. How ?
Thanks and sorry for the formatting.
Best Answer
You can arrange the $10$ letters other than P and S in a row in ${10!\over2}$ ways (divide by $2$ because of the double T). There are $11$ slots between these letters and at the ends of the row. Write P (or S) in slot $k\in[1\>..\>7]$, and $S$ (or P) in slot $k+4\in[5\>..\>11]$. This allows for $2\cdot7$ choices in all. The total number of arrangements therefore comes to $7\cdot10!=25\,401\,600$.