Number of five-letter words with at least one $x$, at least one $y$, and at least one $z$
I had a couple of thoughts about this but none of them fully make sense.
My first thought was get subtract the amount of words with no $x$, $y$, or $z$ from the total amount of words: $26^5 - 23^5$. But then words with only $1$ of them can show up.
My second was subtract words with no $x$, words with no $y$, and words with no $z$ from the total: $26^5-25^5-25^5-25^5$ but that number is less than $0$.
My last was set the first $3$ letters to $a$, $e$, and $s$ then get the last $2$ out of all the letters then account for order with $5!$: $$1 \cdot 1 \cdot 1 \cdot 26 \cdot 26 \cdot 5!$$ But the order only works when all the letters are different.
This problem can be handled with the Inclusion-Exclusion Principle.
There are $26^5$ five-letter words. From these, we must exclude those that do not contain an $x$, a $y$, and a $z$.
There are $25^5$ five-letter words which do not contain an $x$, $25^5$ five-letter words which do not contain a $y$, and $25^5$ words that do not contain a $z$. However, if we subtract $3 \cdot 25^5$ from the total, we will have subtracted too much since we will have subtracted each word which is missing two of the letters twice, once for each way we could have designated one of the missing letters as the missing letter. We only want to subtract these words once, so we must add them to the total.
There are $24^5$ five-letter words which are missing both $x$ and $y$, $24^5$ five-letter words which are missing both $x$ and $z$, and $24^5$ words which are missing both $y$ and $z$. However, if we add $3 \cdot 24^5$ to our total, we will have added too much. This is because we first subtracted each word in which all three of the letters $x, y, z$ are missing three times, once for each way we could have designated one of the missing letters as the missing letter, then added them three times, once for each of the $\binom{3}{2}$ ways we could have designated two of the three missing letters as the pair of missing letters. Thus, we have not subtracted cases in which all three of the letters $x, y, z$ are missing.
There are $23^5$ five-letters words from which the three letters $x, y, z$ are all missing.
Hence, by the Inclusion-Exclusion Principle, the number of five-letter words containing all the letters $x, y, z$ is $$26^5 - \binom{3}{1}25^5 + \binom{3}{2}24^5 - \binom{3}{3}23^5$$
More formally, let $N$ be the number of five-letter words; let $X$ be the number of five-letter words containing an $x$; let $Y$ be the number of five-letter words containing a $y$; let $Z$ be the number of five-letter words containing a $z$. By the Inclusion-Exclusion Principle, we wish to find \begin{align*} |X \cap Y \cap Z| & = N - |X' \cup Y' \cup Z'|\\ & = N - (|X'| + |Y'| + |Z'| - |X' \cap Y'| - |X' \cap Z'| - |Y' \cap Z'| + |X' \cap Y' \cap Z'|)\\ & = N - |X'| - |Y'| - |Z'| + |X' \cap Y'| + |X' \cap Z'| + |Y' \cap Z'| - |X' \cap Y' \cap Z'| \end{align*}
$N$: The number of five-letter words is $26^5$.
$|X'|$: The number of five-letter words that do not contain an $X$ is $25^5$. By symmetry, $$|X'| = |Y'| = |Z'|$$
$|X' \cap Y'|$: The number of five-letter words that contain neither $x$ nor $y$ is $24^5$. By symmetry, $$|X' \cap Y'| = |X' \cap Z'| = |Y' \cap Z'|$$
$|X' \cap Y' \cap Z'|$: The number of five-letter words that do not contain any of the letters $x, y, z$ is $23^5$.
Hence, $$|X \cap Y \cap Z| = 26^5 - \binom{3}{1}25^5 + \binom{3}{2}24^5 - \binom{3}{3}23^5$$