The word can have every letter as long as A,E,I,O,U aren't all in the same word.
Can this be the number of all words of length $10$ without all vowels?
$$ 21^{10}+ \sum_{i=1}^4 {10\choose i} 5^i \cdot 21^{10-i} + \sum_{i=5}^{10} {5\choose 4}{10\choose i} 4^i $$
(all consonant words + all words with less than $4$ vowels + all words with $5$ nonrepeated vowels or more)
Is this correct? If there is a simpler way, I would appreciate.
The number of words in which not all the vowels appear is equal to the number of words in which at least one vowel does not appear.
Let $A, E, I, O, U$ be the set of ten-letter words in which, respectively, the letter $a, e, i, o, u$ does not appear. The set of words in which at least one vowel does not appear is $A \cup E \cup I \cup O \cup U$. By the Inclusion-Exclusion Principle, \begin{align*} & |A \cup E \cup I \cup O \cup U|\\ & = |A| + |E| + |I| + |O| + |U|\\ & \qquad - (|A \cup E| + |A \cup I| + |A \cup O| + |A \cup U| + |E \cup I| + |E \cup O| + |E \cup U| + |I \cup O| + |I \cup U| + |O \cup U|)\\ & \qquad + (|A \cap E \cap I| + |A \cap E \cap O| + |A \cap E \cap U| + |A \cap I \cap O| + |A \cap I \cap U| + |A \cap O \cap U| + |E \cap I \cap O| + |E \cap I \cap U| + |E \cap O \cap U| + |I \cap O \cap U|\\ & \qquad - (|A \cap E \cap I \cap O| + |A \cap E \cap I \cap U| + |A \cap E \cap O \cap U| + |A \cap I \cap O \cap U| + |E \cap I \cap O \cap U|)\\ & \qquad + |A \cap E \cap I \cap O \cap U| \end{align*}
$|A|$: If the vowel $a$ does not appear, then each of the ten positions can be filled in $25$ ways. Thus, $|A| = 25^{10}$.
By symmetry, $$|A| = |E| = |I| = |O| = |U|$$
$|A \cap E|$: If neither of the vowels $a$ nor $e$ appears, then each of the ten positions can be filled in $24$ ways. Thus, $|A \cap E| = 24^{10}$.
By symmetry, $$|A \cap E| = |A \cap I| = |A \cap O| = |A \cap U| = |E \cap I| = |E \cap O| = |E \cap U| = |I \cap O| = |I \cap U| = |O \cap U|$$
$|A \cap E \cap I|$: If none of the vowels $a, e, i$ appears, then each of the ten positions can be filled in $23$ ways. Thus, $|A \cap E \cap I| = 23^{10}$.
By symmetry, $$|A \cap E \cap I| = |A \cap E \cap O| = |A \cap E \cap U| = |A \cap I \cap O| = |A \cap I \cap U| = |A \cap O \cap U| = |E \cap I \cap O| = |E \cap I \cap U| = |E \cap O \cap U| = |I \cap O \cap U|$$
$|A \cap E \cap I \cap O|$: If none of the vowels $a, e, i, o$ appears, there are $22$ ways to fill each of the ten positions. Hence, $|A \cap E \cap I \cap O| = 22^{10}$.
By symmetry, $$|A \cap E \cap I \cap O| = |A \cap E \cap I \cap U| = |A \cap E \cap O \cap U| = |A \cap I \cap O \cap U| = |E \cap I \cap O \cap U|$$
$|A \cap E \cap I \cap O \cap U|$: If none of the vowels $a, e, i, o, u$ appears, there are $21$ ways to fill each of the ten positions. Hence, $|A \cap E \cap I \cap O \cap U| = 21^{10}$.
Thus, by the Inclusion-Exclusion Principle, the number of ten-letter words in which at least one vowel does not appear is \begin{align*} |A \cup E \cup I \cap O \cap U| & = 5 \cdot 25^{10} - 10 \cdot 24^{10} + 10 \cdot 23^{10} - 5 \cdot 22^{10} + 21^{10}\\ & = \binom{5}{1}25^{10} - \binom{5}{2}24^{10} + \binom{5}{3}23^{10} - \binom{5}{4}22^{10} + \binom{5}{5}21^{10}\\ & = \sum_{k = 1}^{5} (-1)^{k - 1}\binom{5}{k}(26 - k)^{10} \end{align*} where $\binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ vowels and $(26 - k)^{10}$ is the number of ways to form a ten-letter word with the remaining $26 - k$ letters of the alphabet.