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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.

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    $\begingroup$ Multiply by the number of ways to arrange P and S themselves in different orders. $\endgroup$ Apr 24 '16 at 15:45
  • $\begingroup$ You mean 12700800 * 14 ? $\endgroup$
    – user312097
    Apr 24 '16 at 15:56
  • $\begingroup$ No, there are only $2!$ ways for arranging $2$ different letters! $\endgroup$ Apr 24 '16 at 16:15
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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 $7\cdot10!=25\,401\,600$.

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  • $\begingroup$ Amazing solution! $\endgroup$
    – 666User666
    Feb 8 at 12:32
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Who said P had to come before S?

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  • $\begingroup$ doesn't matter if P is first or S. $\endgroup$
    – user312097
    Apr 24 '16 at 16:03
  • $\begingroup$ Think about it. $\endgroup$
    – Ben G.
    Apr 24 '16 at 16:03
  • $\begingroup$ No it doesn't matter. $\endgroup$
    – user312097
    Apr 24 '16 at 16:07
  • $\begingroup$ If there are 12700800 permutations where P precedes S, and 12700800 permutations where S precedes P... how many permutations are there in total? $\endgroup$
    – Ben G.
    Apr 24 '16 at 16:08
  • $\begingroup$ No there are 12700800 permutations in which there are 4 letters between P and S. $\endgroup$
    – user312097
    Apr 24 '16 at 16:09

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