This question comes from a completed, marked, and returned exam. It will not likely be reused.


As stated in the question above


First, I note that there are $\binom{10}{5}$ possible groupings.

Second, I note that, if all $4$ females are in the same group, then the remaining fifth member is one of the boys: there are $\binom{6}{1} = 6$ ways to choose the fifth member.

So I conclude $\Pr = \frac{6}{\binom{10}{5}} = \frac{1}{42}$.


I was marked incorrect: the given answer is $\frac{1}{21}$, or exactly twice my answer.

What reasoning led to this conclusion? Why does it seem like some sort of symmetry argument allows us to conclude there are $12$ ways to choose the fifth member?

  • $\begingroup$ Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. $\endgroup$ – N. F. Taussig Feb 28 '19 at 16:55
  • $\begingroup$ @N.F.Taussig apologies i was looking at the answers on the question below as i typed $\endgroup$ – D. Ben Knoble Feb 28 '19 at 17:01
  • $\begingroup$ Another way to see that there are only $126$ possible groups is to observe that if Eloise is one of the four girls, then there are $\binom{9}{4}$ ways to select which four of the other members are in her group. $\endgroup$ – N. F. Taussig Feb 28 '19 at 17:28

You've double-counted the groupings: $\{A, B, C, D, E\}$ and $\{F, G, H, I, J\}$ is the same grouping as $\{F, G, H, I, J\}$ and $\{A, B, C, D, E\}$. Accounting for this double-count, there are $\frac{1}{2} \binom{10}{5}$ distinct groupings. (This is probably the most-common counting mistake of all time. Everyone makes it at least once.)

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  • 2
    $\begingroup$ Those who make it only once don't do much math. $\endgroup$ – saulspatz Feb 28 '19 at 17:07
  • $\begingroup$ @saulspatz lol true. $\endgroup$ – Randall Feb 28 '19 at 17:19

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