Combinatorics involving picking a specific amount of consonants without repeat and vowels (can repeat)

How many words of $13$ letters of the English Alphabet can we constructed which contain $4$ vowels and $9$ different consonants?

This is what I did:
Choose the consonants (diff.) in $\binom{21}9$ ways.
Arrange these consonants in $9!$ ways.
Now we have $4$ empty spots remaining for the vowels, which can be chosen in $5^4$ ways.

My question is: Is this the correct way to think about it?
I'm unsure whether the $5^4$ arranges the vowels or not,

What you calculated is all the different arrangements of $9$ consonants and $4$ vowels, and you did this correctly, but you forgot to arrange the vowels among the consonants, so you need to multiply all of this by ${13 \choose 4}$
• What is the role of $5^4$ here then? Because I thought that $5^4$ would arrange these vowels as well. – OneGapLater Oct 17 '17 at 2:33
• @ActuarialStudent101 Yes and no. The $5^4$ is for picking $4$ vowels in some order among themselves (i.e a 'first', 'second', 'third', and 'fourth' vowel, but you now need to place those 4 vowels amodst the 9 consonants, i.e pick out 4 positions out of the 13 possible positions for those vowels to go relative to the whole word. – Bram28 Oct 17 '17 at 12:32
You've got 13 letters. Choose 9 of them to make your consonants. Then you've got 21 consonants for the first, 20 for second, etc. Then 5 vowels to put in each of the other spots. So $\binom{13}{9}\cdot 21^{\underline{9}} \cdot 5^4$
• This answer uses the falling factorial notation $x^{\underline{n}}$, which is defined by $$x^{\underline{n}} = x(x - 1)(x - 2) \ldots (x - n + 1) = \prod_{k = 0}^{n - 1} (x - k)$$ Therefore, $$21^{\underline{9}} = 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 = \frac{21!}{12!} = \frac{21!}{12!9!} \cdot 9! = \binom{21}{9}9!$$ – N. F. Taussig Oct 17 '17 at 10:19