I wrote a probability question I can't solve $\newcommand{\nCk}[2]{{}^{#1}C_{#2}}$
I wrote this question to prep my students for their midterm, and I realized when I sat down to solve it that I can't figure out the right way to think about it.
A costume shop has $7$ costumes available to rent, $4$ of each. You and $9$ friends go to the costume shop independently and each pick a costume. What is the probability you arrive at the party and there are exactly $5$ distinct costumes? (You are the only 10 guests)
I know from simulation that the solution should be something like $0.232402$, perhaps it is $\nCk{7}{5}\cdot 48\cdot\frac{\nCk{15}{5}}{\nCk{28}{10}}$. I can justify the $\nCk{7}{5}$ and the $\nCk{15}{5}$ in the numerator, but not the $48$. $\nCk{7}{5}$ because of the $7$ costumes $5$ are chosen. We need to ensure that $1$ of each are worn by $5$ guests, but the other $5$ are free to choose from the $15$ (hence $\nCk{15}{5}$). But I'm sure I'm thinking about this not quite right. I'd love any explanations. Thanks!

I find it easier to think of this in terms of cards. Suppose I have a partial deck of cards with all 4 aces, all 4 twos, and so on, up until all 4 sevens. I shuffle these 28 cards and then look at the top 10. What is the probability that exactly 5 ranks are presents among these 10 cards.
The denominator is $\binom{28}{10}$. For the numerator, we must first choose the 5 ranks that will be present, giving a factor of $\binom{7}{5}$. Once these are chosen, we must choose the partition of 10, as Alan noted, that these cards will form. From Alan's answer, they are
43111, 42211, 33211, 32221, 22222
We first count the number of ways of selecting cards according to the first partition. We must choose the rank that will have 4 cards, the rank that will have 3, and the three ranks that will have 1. There are
$$
\binom{5}{1,1,3} = \frac{5!}{1!1!3!} = 20
$$
ways to do that. Once that selection is made, we must actually choose the cards. There are
$$
\binom{4}{4}\binom{4}{3}\binom{4}{1}^3
$$
ways to do that. Continuing with this reasoning for the other four partitions, our final answer will be
$$
\frac{\binom{7}{5}}{\binom{28}{10}}\left(
\binom{5}{1,1,3}\binom{4}{4}\binom{4}{3}\binom{4}{1}^3
+ \binom{5}{1,2,2}\binom{4}{4}\binom{4}{2}^2\binom{4}{1}^2
+ \binom{5}{2,1,2}\binom{4}{3}^2\binom{4}{2}\binom{4}{1}^2
+ \binom{5}{1,3,1}\binom{4}{3}\binom{4}{2}^3\binom{4}{1}
+ \binom{5}{5}\binom{4}{2}^5
\right).
$$
If my calculations are correct, this should be
$$
\frac{6608}{28405} \approx 0.232635.
$$
