# arrangement of the word $\bf{PERMUTATION}$ in which exactly $4$ letters in b/w $\bf{P}$ and $\bf{N}$

Total no. of arrangement of the word $\bf{PERMUTATION}$ in which there is there are exactly

$4$ letters in between $\bf{P}$ and $\bf{N}$.

$\underline{\bf{My\; Trial \; solution}}::$ Here we will form different cases:

$\bullet \;$ If $\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

$\bullet \;$ If $\boxed{-}\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

$\bullet \;$ If $\boxed{-}\boxed{-}\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\boxed{-}\boxed{-}\boxed{-}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

$\bullet \;$ If $\boxed{-}\boxed{-}\boxed{-}\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\boxed{-}\boxed{-}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

$\bullet \;$ If $\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\boxed{-}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

$\bullet \;$ If $\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}\;$ Then no. of ways $\displaystyle = \frac{9!}{2!}\times 2! = 9!$

So Total no. of ways $= 9!+9!+9!+9!+9!+9!=6\times 9!$

But the above method is very lengthy, can we solve it any short and better method,

Thanks

• What you wrote is in principle quick. We assume $P$ and $N$ can be permuted. The leftmost of these can occupy one of $6$ positions. There are $2$ ways to decide which of P or N will occupy that position. For each choice, the remaining slots can be filled in $\frac{9!}{2!}$ ways. Multiply. – André Nicolas Oct 1 '14 at 5:35

Fix P and N so that P is the furthest to the left. Notice that P can occupy only 6 possible positions. Now ignore P and N, and permute everything else, done in $\frac{9!}{2!}$ ways. P and N can switch, so we multiply by a factor of $2!$.

therefore, $6*\frac{9!}{2!}*2!=6*9!$

Treating $\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}$ as a single object, there are $^9 P_4$ ways to pick and arrange other letters of the word to put inside the spaces. Then arrange it alongside the remaining letters, so we get $^9\!P_4\times (11-6+1)! = ^9\!P_4 \times 6! = 2177280$ total arrangements, which is the same as $6\times 9!$

EDIT: @user84413 has pointed out that I should account for the repetition of T and interchangeability of P,N. This can be done by case studies, where we consider how many "T"s go inside the $\boxed{P}\boxed{-}\boxed{-}\boxed{-}\boxed{-}\boxed{N}$ object.

When there are no T's inside, we select the letters to go inside out of $\left\{ E,R,M,U,A,I,O \right\}$. This gives $^7\!P_4\times \frac{(11-6+1)!}{2!}$ arrangements.

When there is 1 T inside, we select the others letters to go inside out of the same set. This gives $^7\!C_3\times 4!\times(11-6+1)!$ arrangements.

When both T's are inside there are similarly $^7\!C_4 \times \frac{4!}{2!}\times(11-6+1)!$ arrangements.

Multiplying their sum by $2!$ to account for interchangeability should give the same result i.e. $\displaystyle 2! \times (^7\!P_4\times \frac{6!}{2!} +^7\!C_3 \times 4! \times 6! + ^7\!C_4 \times \frac{4!}{2!}\times 6! ) = 2177280$.

• This gives the right answer, but I believe you want to take into account that there are 2 letters the same and that P and N can be interchanged. – user84413 Oct 1 '14 at 23:15