The probability of random permutation of $POSSESSION$ without two adjacent S's appearing twice and next to each other Consider a random permutation of the letters in the English word
POSSESSION.
Is the probability that exactly two adjacent S’s, SS in other
words, appear twice and the two SS’s are not next to each other $\frac{73080}{75600}$?
The following is my solution:
total combination: $\frac{10\,!}{4\,! 2\,!}=75600$(due to there are 4 S's and 2 O's in this permutation)
two SS’s appear twice and are next to each other: $\frac{7\,!}{2\,!}=2520$
$\;\boxed{SSSS}\ P\ O\ E\ I\ O\ N$
$P(two\ SS’s\ appear\ twice\ and\ the\ two\ SS’s\ are\ not\ next\ to\ each\ other)=1-\frac{2520}{75600}=\frac{73080}{75600}$.
However, I am not sure whether I am wrong.
I think this probability is so large that I am confused.
Is there anything wrong in my solution?
 A: Make two blocks of $\;\boxed{SS},\;\boxed{SS}$
and permute the remaining letters  $POEION$ in $\frac{6!}{2!}$ ways
The two boxes can now be inserted in the $7$ gaps (including ends) between the other letters in $\dbinom72$ ways,
thus $Pr = {\dfrac{6!}{2!}\dbinom72}{\Large/}{{\dfrac{10!}{2!4!}}}$
A: Alternative approach:
To calculate the denominator, I would reason that you have $4$ identical S's, $2$ identical O's, and $4$ other distinct letters.
So, my computation of the denominator is
$$\binom{10}{4} \times \binom{6}{2} \times 4! = \frac{(10)!}{(4!)(2!)},$$
which agrees with the OP's (i.e. original poster's) calculation.
For the numerator:
You can calculate all of the ways of having $2$ pairs of S's, and then deduct all of the ways of having one block of $4$ S's.

For $2$ pairs of S's, you have $8$ units to permute.  $2$ of the units are the identical pairs of S's, and $2$ of the units are the identical letters, the O's.
So, the computation here is
$$\binom{8}{2} \times \binom{6}{2} \times 4! = 10080.\tag1 $$

For one unit of $4$ S's, and 6 other letters to permute, two of which are O's, you have
$$\binom{7}{2} \times 5! = 2520.\tag2 $$
Taking the difference of (1) and (2) above gives
$$10080 - 2520 = 7560 = ~\text{the numerator}.$$
