Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Note: there are 33 distinct men.
The problem is similar to this one:
How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was:
$$
\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}
$$
But there are clearly a lot of solutions overlapping.
Best Answer
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
First we’ll pick the team containing the youngest of the $33$ men; there are $\binom{32}{10}$ ways to do that, since we need only choose his $10$ teammates. Then we choose the team containing the youngest of the remaining $22$ men; this can be done in $\binom{21}{10}$ ways. That leaves $11$ men to form the third team. This approach does not overcount: there’s always one team that contains the youngest man and one that contains the youngest man not on that first team. Thus, there are
$$\binom{32}{10}\binom{21}{10}\tag{2}$$
ways to choose the three teams.
Of course it would be a good idea to make sure that $(1)$ and $(2)$ actually yield the same result:
$$\begin{align*} \frac16\binom{33}{11}\binom{22}{11}&=\frac16\cdot\frac{33!}{11!22!}\cdot\frac{22!}{11!11!}\\\\ &=\frac13\cdot\frac{33\cdot32!}{11!22!}\cdot\frac12\cdot\frac{22\cdot 21!}{11!11!}\\\\ &=\frac{11\cdot32!}{11!22!}\cdot\frac{11\cdot21!}{11!11!}\\\\ &=\frac{32!}{10!22!}\cdot\frac{21!}{10!11!}\\\\ &=\binom{32}{10}\binom{21}{10}\;. \end{align*}$$