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How many words of $13$ letters can be constructed from the English alphabet which contain $4$ vowels and $9$ different consonants (vowels can be the same).
This is what I think:
Pick the four vowels (does this arrange them as well?) = $5^4$.
Choose the consonants = $\binom{13}9$
Now then arrange these consonants in $9!$ ways
Total = $5^4 \binom{13}9 9!$.

My friend got this:
$5^4 \binom{21}9 \binom{13}4$ to arrange the four vowels (with the last binomial).

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marked as duplicate by N. F. Taussig combinatorics Oct 17 '17 at 2:37

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Both you and your friend made mistakes.

Choose four of the $13$ positions for the vowels: $\binom{13}{4}$

Since vowels may be repeated, each of these four positions can be filled in $5$ ways: $5^4$

Choose which nine of the $21$ consonants will fill the remaining positions: $\binom{21}{9}$

Arrange the chosen consonants in those positions: $9!$

Hence, the number of permissible words is $$\binom{13}{4}5^4\binom{21}{9}9!$$

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