These are two problems from my combinatorics assignment that I'm not quite confident in my answer. Am I thinking of these the right way?
Problem 1: On rolling 16 dice. How many of the $6^{16}$ possible (ordered) outcomes would contain exactly three quadruplets? (the same number appearing four times.)
Answer: There are $ \binom{16}{4, 4, 4, 1, 1, 1, 1} $ ways to choose the dice (four quadruplets and four singlets).
There are $\binom{6}{3}$ possible values the quadruplets can have.$^1$
The possible values of the singlets can repeat, yielding $3^4$ in total. But three of those combinations would be such that they all have the same value; and we can't have another quadruplet. So the number of possible values for the singlets must be $3^4 - 3$.
Thus, the total number of possible values is
\begin{equation*} \binom{16}{4,4,4,1,1,1,1} \binom{6}{3} (3^4 - 3) = 2 \, 361 \, 078 \, 720 \, 000 \end{equation*}
- I think this should be changed to $\frac{6!}{3!}$ instead of $\binom{6}{3}$?
problem: Four friends and 21 boys are to be randomly divided into five teams of 5 players. How many of the $\binom{25}{5,5,5,5,5} \frac{1}{5!}$ possible ways of creating five teams will have each of the four friends play for a different team?
Answer: There are $\frac{4!}{4!}$ ways to distribute the four friends into four teams. There are $\binom{21}{1}$ ways to choose a leader for the fifth team from the remaining $21$ kids.
There are $\binom{20}{4,4,4,4,4} \frac{1}{5!}$ ways to distribute the remaining $20$ kids among the five teams.
This yields a total number of possible teams:
\begin{equation*} \frac{4!}{4!} \binom{21}{1} \binom{20}{4,4,4,4,4} \frac{1}{5!} = 3 \, 547 \, 982 \, 095 \, 257 \, 600 \end{equation*}