<s>[INST] Two combinatorics problems. I'm not 100% confident in my answers 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*}

 [/INST] The following is an analysis of the second problem.  It yields a somewhat different answer.  Added: It is followed by an analysis of the first problem, again yielding a somewhat different answer.
Let the four friends be called A, B, C, and D. Of these four, A is the leader, her friends always let her go first.  And B is second in status. Then comes C, with poor D always last.  Sad, you may think, but irrelevant. However, it will turn out that the pecking order of the four friends is useful for correct counting. (The character analysis is borrowed from Stephen Leacock's essay "A, B, and C.")
First the coach chooses the $5$ people who will not have one of the friends on their team.  This can be done in $\binom{21}{5}$ ways.  
Now A picks from the remaining $16$ people the $4$ people who will be on her team.  This can be done in $\binom{16}{4}$ ways.  Once A is through choosing, it is B's turn to choose.  For each choice that A made, B has $\binom{12}{4}$ choices. Continue. We get that the number of choices is
$$\binom{21}{5}\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}.$$
This can be simplified to
$$\frac{21!}{5!4!4!4!4!},$$
and then we could compute.
Added: I had not looked at the first problem, since there seemed to be agreement about the answer. But here is an analysis.  I take it that $3$ quadruples means that there are $3$ numbers that occur exactly $4$ times each (but not necessarily in a row), and no other number occurs $4$ times.  (The other two analyses also make this assumption.)
The $3$ numbers that will appear $4$ times can be chosen in $\binom{6}{3}$ ways.  Now suppose we have done the choosing, and that the chosen numbers, listed in increasing order, are $p$, $q$, and $r$.  Choose the places that the $p$'s will go. This can be done in $\binom{16}{4}$ ways.  For each such choice, there are $\binom{12}{4}$ ways to choose where the $q$'s will go, and for each such choice there are $\binom{8}{4}$ ways to choose where the $r$'s will go.  Once we have done that, there are, as in the OP's analysis, $3^4-3$ ways to fill in the remaining open slots. We get therefore a total of
$$\binom{6}{3}\binom{16}{4}\binom{12}{4}\binom{8}{4}(3^4-3).$$
To compare with the other answers, we simplify a little, to
$$\binom{6}{3}\frac{16!}{4!4!4!4!}(3^4-3).$$
Comment: Maybe multinomial coefficients are slightly more tricky to work with than combinations of binomial coefficients. In the first problem, once the locations of the $3$ quadruples have been determined, all that remains is to fill in the remaining $4$ spots: no choosing of $1$ spot then $1$ spot then $1$ spot then $1$ spot is involved.
</s>