2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Note that $\frac{6}{\binom{10}{5}} = \frac{6}{252} = \frac{1}{42}$. $\endgroup$ – N. F. Taussig Feb 28 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 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 at 17:28
7
$\begingroup$

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.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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