# Combinatorics with vowels and consonants [duplicate]

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).

## marked as duplicate by N. F. Taussig combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 17 '17 at 2:37

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 $$\binom{13}{4}5^4\binom{21}{9}9!$$