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Suppose you have an alphabet of $5$ letters: $\left \{ a,b,c,d,e \right \}$. Let's consider a set of words of length 25. What is the number of words that have $5$ $a$, $5$ $b$, $5$ $c$, $5$ $d$, and $5$ $e$ in it?

This is my approach: There are $\binom{5}{1}$ choices for the first letter, and then $\binom{24}{4}$ choices for placing that letter in the word. Then we have $\binom{4}{1}$ choices for the second letter, and then $\binom{19}{4}$ choices for placing that letter in the word. And we continue so on.

At the end, I find that the total number of such words is equal to $$\binom{5}{1} \binom{24}{4} \:\:\: \binom{4}{1} \binom{19}{4} \:\:\: \binom{3}{1} \binom{14}{4} \:\:\: \binom{2}{1} \binom{9}{4} \:\:\: \binom{1}{1}\binom{4}{4} $$

Is my reasoning correct?

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    Did you mean set of words of length 25? – 2017-02-20
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    @Smylic Exactly, made an edit! Thanks. – 2017-02-20
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    You solution is correct with an exception of $\binom{14}{1}$, $\binom{9}{1}$ and $\binom{4}{1}$ that looks like a typo. – 2017-02-20
  • 0
    @Smylic Yes, those are indeed typos, my bad. – 2017-02-20

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You solution is correct. Here I'll give another one. We have $\binom{25}{5}$ sets of positions of letter $a$, $\binom{20}{5}$ sets of positions of letter $b$ etc. So the answer is $$\frac{25!}{5!^5} = \binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5} = 5\binom{24}{4}\cdot 4\binom{19}{4} \cdot 3\binom{14}{4} \cdot 2\binom{9}{4} \cdot 1\binom{4}{4}\\ = \binom{5}{1}\binom{24}{4}\cdot \binom{4}{1}\binom{19}{4} \cdot \binom{3}{1}\binom{14}{4} \cdot \binom{2}{1}\binom{9}{4} \cdot \binom{1}{1}\binom{4}{4}$$