2
$\begingroup$

A committee of 6 people is to be chosen from a group consisting of 7 men and 8 women. If the committee must consist of at least 3 women and at least 2 men, how many different committees are possible?

The answer given in the text is $\binom{8}{3} \binom{7}{3} + \binom{8}{4} \binom{7}{2} = 3430$. I understand this line of thinking--first count all the ways we can choose 3 of the 8 women and 3 of the 7 men. Then, add to this all of the we can choose 4 of the 8 women and 2 of the 7 men.

I thought about this question a bit differently, however, and am failing to see where my line of thinking is flawed. This is how I performed the calculation:

$$\binom{8}{3} \binom{7}{2} 10 = 11760$$

In other words, we choose 3 women and then 2 men. Then, we are left with 10 other choices, the gender of whom is irrelevant. What am I doing wrong?

$\endgroup$
2
$\begingroup$

You are counting things too many times: If the sixth member is a female, then you get a committee of $4$ women and $2$ men, but exactly this committee was chosen $3$ other times. If the sixth member is a man then this $3$ women / $3$ men committee was chosen $2$ other times.

Note that you can thus fix your reasoning to compute: $$\binom{8}{3}\binom{7}{2}\left(\frac{5}{4}+\frac{5}{3}\right)=3430.$$

$\endgroup$
  • $\begingroup$ I see it now. To be more clear, if the sixth choice (from the last choice of ten) was a female, I would have a subcommittee of four females. However, this same subcommittee of females is counted four times--once where each of the four is counted as the "last" choice, and the other three are chosen in the "first" $\binom{8}{3}$ choice. $\endgroup$ – Brennon Bortz May 12 '14 at 17:49
  • $\begingroup$ @BrennonBortz: Exactly. $\endgroup$ – DKal May 12 '14 at 17:50
  • $\begingroup$ @BrennonBortz, I've edited to add a computation along the lines of your idea. $\endgroup$ – DKal May 12 '14 at 17:56

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.