There are $11$ footballers in a football team. They will be split into groups, each containing $3$ or $4$ people. How many ways are there to split them that way?

I have obtained that the groups will contain $4$, $4$, and $3$ people.

  • 1
    $\begingroup$ Hint: the first thought, $\binom {11}4\times \binom 74 \times \binom 33$ double counts. For example, you count $\{1,2,3,4\}\{5,6,7,8\}\{9,10,11\}$ again as $\{5,6,7,8\}\{1,2,3,4\}\{9,10,11\}$. You have to account for the symmetry between the two groups of four. $\endgroup$ – lulu Mar 6 '17 at 13:13

I think there's been some confusion in the solutions. The way I read the question there is no difference between the two groups of $4$. That is to say, the division $\{1,2,3,4\}\{5,6,7,8\}\{9,10,11\}$ coincides with the division $\{5,6,7,8\}\{1,2,3,4\}\{9,10,11\}$.

The answer $\binom {11}4\times \binom 74\times \binom 33$ supposes that the two groups of four can be distinguished. This would be the correct answer if we had been asked "How many ways are there to choose four players for the Red Team, four for the Blue Team, and three for the Yellow Team?". With the question as it is written, I think the better answer is $$\frac 12\times \binom {11}4\times \binom 74\times \binom 33=\boxed {5775}$$


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