If there are $20$ students in a class, in how many ways can a professor give out $4$ A's, $3$ B's ,$4$ C's, and $9$ F's?
What I did was use the binomial theorem:
$\frac{20!}{4!\cdot 3!\cdot4! \cdot9!} = 1,939,938,000$
Is this a correct way of approaching a problem like this?
Best Answer
Yes.
If you want to convince yourself, beyond a hunch that you have remembered the right formula, place the $20$ students into order - $20!$ ways of doing that. Then give the first four an A, the next three a B, and so on.
Because the four A students could come in any order, you have over-counted by a factor of $4!$. Because the three B students could come in any order, you have over-counted by a factor of $3!$. And so on.
Formulae are all very well, but they all look the same sometimes, so it’s good to be able to work the answer out for yourself.