The proposed problem was the following (translated into English by me):
"A school has $16$ students interested in participating in a team competition. First, they must split themselves into $8$ pairs. Then, each of those pairs must choose (among themselves) someone to be their leader. Finally, two teams of $8$ students are formed by choosing $4$ pairs to compose the first team, which already locks the opposing team.
Considering two teams to be equal if, and only if, they are composed of the same pairs (with pairs being equal if they are composed of the same students, with the same leader), in how many distinct ways can these $16$ students divide themselves into two teams of eight?"
Solution Provided (translated by me):
First, compute the total amount of distinct pairs that can exist. This is clearly given by $C_{16,2} = \frac{16!}{14!2!} = 15\cdot8$. Since each pair has two choices of leader, we also multiply by two, to obtain $16\cdot15 = 240$
Now, multiply the amount of distinct pairs by the total number of ways of choosing $4$ out of the $8$ formed pairs to obtain the final answer:
$C_{8,4} \cdot 240 = \frac{8!}{4!4!} \cdot 240 = 70 \cdot 240 = 16800$
My issues:
I don't believe that using $C_{16,2}$ makes any sense, as (in my mind) after you choose the first pair, you cannot choose any other pair that contains either of the students in the first one you chose (and so on).
I also don't believe this number correctly represents the total amount of ways of forming 8 pairs under these conditions, which is the number that would make sense to use in place of $C_{16,2}$ (in my opinion).
My attempt:
We begin by computing the number of ways of constructing $8$ distinct pairs out of $16$ students.
For the first pair, we have $16 \choose 2$ possible ways. For the second pair, $14\choose 2$. And so on until the final pair which is $2 \choose 2$. So the total number of ways of assembling 8 pairs out of 16 students seems to be given by:
$\prod_{n=1}^{8}{2n \choose 2}$
Which works out quite nicely to $\frac{16!}{2^{8}}$
But this is not quite the number I am looking for yet. Since I don't actually care about the order of the pairs, I have to divide by $8!$, since when multiplying them out like I did above, I consider choosing some pair $A$ before some pair $B$ different then choosing first $B$ and then $A$, when it clearly is not.
So, the final number works out to: $\frac{16!}{8!2^{8}}$.
Like it was done in the official solution, I must now multiply by $2^{8}$, since each individual has two choices of leaders. This gives me the total number of ways of forming $8$ distinct pairs with leaders which is $\frac{16!}{8!}$
The conclusion is now the same as in the official solution: We now multiply by the amount of different ways of choosing $4$ out of the $8$ pairs. This gives:
$\frac{16!}{8!}\cdot C_{8,4} = \frac{16!}{8!}\cdot\frac{8!}{4!4!} = \frac{16!}{4!4!}$
And finally, we must divide by two since choosing, for example, pairs $A,B,C,D$ (out of pairs $A,B,C,D,E,F,G$) or choosing $D,E,F,G$ makes no difference on the resulting two teams.
So my final answer turns out to be $\frac{16!}{4!\cdot 4!\cdot 2}$, which is not even close to the number obtained in the official one.
The question: Which one (if any) is right, and why is the other one (or both) wrong?
Best Answer
Your solution and answer look fine to me. Here's an alternate approach.
We line up the $16$ students. The first $8$ will form one team and the next eight another. In each team, the first two form one pair, the next two another, and so on. In each pair, first student is the leader.
Now, the $16$ students can be arranged in a row in $16!$ ways. But we don't care about the order of the $4$ pairs inside a team. So, divide by $4! \cdot 4!$ for the two teams. Also, the order of the teams doesn't matter. So divide by $2$.
So, the answer is $\dfrac{16!}{4! \cdot 4!\cdot 2}$