Arrangements of MISSISSIPPI with all S's and P's separated

If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $k\frac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=k\frac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks

Leaving out the $S's$ for the moment, there are $\frac{7!}{4!2!} = 105$ permutations of $PPMIIII$

Of these $\frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.

For the together cases, we need to borrow an $S$ to separate them, e.g. $\boxed{PSP}MIIII$

The remaining $3\; S's$ can be inserted in the gaps between units (including ends) in $\binom73$ ways

For cases with $P's$ already separate, e.g. PMPIIII, the $4\; S's$ can be inserted similarly in $\binom84$ ways.

Thus permissible permutations $= 30\binom73 + 75\binom84 = 6300$

For better or worse, this give a value of $k=1$

Use of heavy artillery

To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.

Define polynomials for $k\geq 1$ by $q_k(x) = \sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$.

e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$

For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$

The number of permutations will be given by

$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$

Wolframalpha gives the answer as $151,200$

Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$