Letter permutations in 'COOLEEMEE' 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:)
 A: 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.
A: 
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?

If you consider 8 spaces (7 letters + 1) here, it means that you are adding the two Os as a unique block, so you should obtain $\frac{7!}{4!}\binom{8}{1}=\frac{7!}{4!}\cdot 8$, which are exactly the cases that you have to remove from the total.
This is basically the first step of the inclusion-exclusion principle: you're considering all the possible permutations and then you exclude the ones that have consecutive Os.
Otherwise if you consider 9 spaces, you have $\frac{7!}{4!}\binom{9}{2}=\frac{9!}{4!2!}$, which is basically splitting the initial problem into two separate steps.
The problem with the letter E is a bit more complicated instead, I can't guarantee this is the easiest solution:
Your goal is to remove once from the total every word that has at least two consecutive Es
you can start by removing the permutations of the block [EE] and the letters "COOLEEM". But then you notice that you also removed:

*

*twice (one too much) the sequences that contain two couples of E (..EE...[EE]...,...[EE]...EE...)


*twice (one too much) the sequences that contain at least three consecutive E (...E[EE]...,...[EE]E...)


*thrice (two too many) the sequences that contain at least four consecutive E (...EE[EE]...,...E[EE]E...,...[EE]EE...).
You correct the first error by adding once the permutations of the symbols "[EE][EE]COOLM".
You try to correct the second error by adding once the permutations of the block [EEE] and the letters "COOLEM", but then again you notice that you're adding twice the words with 4 E (...E[EEE]...,...[EEE]E) thus amending your second and third error.
This is also somehow a non-canonical application of the inclusion-exclusion principle.
A: Here we are looking for the number of words with no consecutive equal characters at all. Such words are called Smirnov or Carlitz words. See example III.24 Smirnov words in Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
A generating function for the number of Smirnov words over an $n$-ary alphabet is given by
\begin{align*}
\left(1-\frac{nz}{1+z}\right)^{-1}\tag{1}
\end{align*}

Here we consider an alphabet $\mathcal{V}=\{C,E,L,M,O\}$ with $n=5$ letters. Using $[z^k]$ to denote the coefficient of $z^k$ of a series we calculate
\begin{align*}
&\color{blue}{[CE^4LMO^2]\left(1-\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^{-1}}\tag{2}\\
&\qquad=[CE^4LMO^2]\sum_{j=0}^{\infty}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\tag{3}\\
&\qquad=[CE^4LMO^2]\sum_{j=5}^{9}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\tag{4}\\
&\qquad=[CE^4LMO^2]\sum_{j=5}^{9}\left(C+E\left(1-E+E^2-E^3\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad\left.+L+M+O(1-O)\right)^j\tag{5}\\
&\qquad=120-1\,440+6\,300-11\,760+7\,560\tag{6}\\
&\,\,\color{blue}{\qquad=780}
\end{align*}
in accordance with @trueblueanils answer.

Comment:

*

*In (2) we use the generating function (1). Since we want to identify specific powers of letters, we use a term $\frac{a}{1+a}$ for each letter in $\mathcal{V}$.


*In (3) we expand the geometric series.


*In (4) we observe that each term $\frac{a}{1+a}=a\left(1-a+a^2-\cdots\right)$ contains at least a factor $a$ so that the index range of of the series can be restricted to $j=5,\ldots,9$.


*In (5) we expand $\frac{a}{1+a}$ for each of the letters in $\mathcal{V}$. We skip all powers of letters, which do not contribute to $C,E^4,L,M,O^2$.


*In (6) we calculate the coefficients of the powers of $j=5,\ldots,9$ with some help of Wolfram Alpha.
