How many words of length $10$ are there in which not all of the vowels A, E, I, O, U appear?

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.