# In how many ways can 3 distinct teams of 11 players be formed with 33 men?

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.

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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*}

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And yet another approach...

Line the 33 men up, and assign the first 11 to Team A, the second 11 to Team B and the remaining 11 to Team C.

The line can be formed in $33!$ ways. But this counts as distinct the $11!$ rearrangements of the 11 men within each of the three teams.

Thus, we have $33! / (11!)^3$ ways to set up the three teams.

This calculation counts as distinct having a particular group of 11 on Team A, or on Team B or on Team C. If this is not the case (I'm not convinced!) then divide by $3!$

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