I'm considering the word 'COOLEEMEE'. The first part is asking the number of distinguishable permutations of these 9 letters, which should be $9!/(2!4!)=7560$.
The next part is asking how many distinguishable permutations of these 9 letters are there if consecutive O’s are forbidden?
(And the last part if how many distinguishable permutations if consecutive E’s are forbidden)
If we don't allow consecutive O's, then I think we could first permute the rest letters 'CLEEMEE', so there're $7!/4!$ ways. Then there're 8 spaces and I want to insert 2 O's, so should that be $\frac{7!}{4!}\binom{8}2$ in total?
Then similarly if consecutive E’s are forbidden, there're $\frac{5!}{2!}\binom{6}4$ total ways. Is my reasoning correct? I feel like I missed some cases but I'm not sure. I've seen some similar examples which used the inclusion-exclusion principle, but I don't know if I need to use that here. Thanks:)
Best Answer
It is best to first find arrangements with no two $E's$ together, inserting the $E's$ in the dashes ($-$)
$-C-O-O-L-M-\;\;:\binom64\frac{5!}{2!} = 900$
From this, we need to subtract arrangements with the two $O's$ together,
$-C-[OO]-L-M-$, viz $\binom54 4! = 120$,
giving a final answer of $\;900-120=780$
Added to address OP's query
I have taken the problem to mean that neither $E's\;nor\;O's$ should be together since you have mentioned inclusion-exclusion. If they are two separate sub-problems, one with no $E's$ together, another with no $O's$ together, you already know the procedure.