Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
up vote 2 down vote accepted

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


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


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:

$$\begin{align*} \frac16\binom{33}{11}\binom{22}{11}&=\frac16\cdot\frac{33!}{11!22!}\cdot\frac{22!}{11!11!}\\\\ &=\frac13\cdot\frac{33\cdot32!}{11!22!}\cdot\frac12\cdot\frac{22\cdot 21!}{11!11!}\\\\ &=\frac{11\cdot32!}{11!22!}\cdot\frac{11\cdot21!}{11!11!}\\\\ &=\frac{32!}{10!22!}\cdot\frac{21!}{10!11!}\\\\ &=\binom{32}{10}\binom{21}{10}\;. \end{align*}$$

share|cite|improve this answer

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 $3!$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.