$10$ people ($6$ male, $4$ female) divided into $2$ equal groups: what is the probability that all females are in the same group?

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

Problem

As stated in the question above

Work

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

Question

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?

• Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. – N. F. Taussig Feb 28 at 16:55
• @N.F.Taussig apologies i was looking at the answers on the question below as i typed – D. Ben Knoble Feb 28 at 17:01
• 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. – N. F. Taussig Feb 28 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.)