The prompt asks you to find the number of k member teams that can be created from n people.
The number of teams that can be created depends on the number of people left after creating the first group, considering there are 10 people, and we make teams of 4, we have 2 teams and 2 people get left out, so we must only choose so many members in a team that all n people can be accommodated.
The way I went on about solving the problem was to put one person in each team to begin with, since we can't have a team with no players, so we have $n\choose k$. Now we are left with n – k people, since number of teams is equal to number of players already in a team, now we can do something like $${n\choose k}(n-k)^k$$ to divide rest of the n-k players into k teams. Is this the correct way to approach this problem?
What I don't understand is that this method considers we already know how many teams there are, how would we solve if we don't know how many teams exist. Is this similar to finding all possible binary sequences with n zeros and k ones, since we don't know how many sequences exist, but we know what goes in those sequences. Applying that formula here, $${n + 1 \choose k} $$ when $k \leq n+1$ and 0 when $k > n + 1$
Best Answer
Each team must have $k$ players, so $k$ divides $n$. Then can you see why
$$\binom{n}{n-k} \binom{n-k}{n-2k} \binom{n-2k}{n-3k} \ldots$$ works? At the first stage, we pick $n-k$ people to stay behind, and the rest form a team. In the next stage, we pick $n-2k$ people to stay behind, and the rest form a new team, and so on.
Now, using the suggestive notation I used above, let's simplify:
$$\frac{n!}{(n-k)!k!} \cdot \frac{(n-k)!}{(n-2k)!k!} \frac{(n-2k)!}{(n-3k)!k!} \ldots = \frac{(mk)!}{k!^{m}}$$
where $m = n/k$.