I can't find where I am overcounting in the problem
How many 10-letter words do not contain all the vowels.
What I do is to count all the words that have at least 5 different vowels:
$$ \binom{10}{5}5!*26^5, $$
this means that I have the $5$ different vowels which can be fit in $10$ spaces. The rest of the word could be also a 5-letter word (no constriction imposed).
So the result should be:
$$ 26^{10} - \binom{10}{5}5!*26^5. $$
What is wrong with it?
The question of "what is wrong with your count" has been answered above in the comments by @BrianM.Scott. Here I begin the setup for the intended solution of the problem.
Let $N_a$ denote the event that there are no $a$'s in the word. Let $N_e$ denote the event that there are no $e$'s in the word. Etc. for the rest of the vowels.
Let our universal set, $S$, be all words that are of length ten (regardless of number of vowels present).
We have then $N_a^c$ represents the words of length ten which do have at least one $a$.
The number we are asked to calculate is $|N_a^c\cap N_e^c\cap N_i^c\cap N_o^c\cap N_u^c|$, in other words, the number of words which have at least one of each vowel.
Simplifying using DeMorgan's laws and inclusion exclusion, we have:
$\begin{array}{rl}|N_a^c\cap N_e^c\cap N_i^c\cap N_o^c\cap N_u^c| &= |S\setminus (N_a\cup N_e\cup N_i\cup N_o\cup N_u)| \\ &= |S| - |N_a\cup N_e\cup N_i\cup N_o\cup N_u|\\ &=|S|-|N_a|-|N_e|-|N_i|-\dots+|N_a\cap N_e|+\dots\\ &~~~~~~~~~~~~-|N_a\cap N_e\cap N_i|-\dots\end{array}$
Notice that $|N_a|=|N_e|=|N_i|=\dots$ and that $|N_a\cap N_e|=|N_a\cap N_i|=\dots$ etc. to simplify the above expression to:
$|S|-\binom{5}{1}|N_a|+\binom{5}{2}|N_a\cap N_e|-\binom{5}{3}|N_a\cap N_e\cap N_i|+\binom{5}{4}|N_a\cap N_e\cap N_i\cap N_o|-\binom{5}{5}|N_a\cap N_e\cap N_i\cap N_o\cap N_u|$
How big is something like $|N_a\cap N_e|$?
These are the words which simultaneously don't have $a$'s and don't have $e$'s.
Simplify all of the above terms similarly.