Seven women and nine men are on the faculty in the mathematics department at a school. $\\$ How many ways are there to select a committee of five members of the department if at least one woman and at least one man must be on the committee?

I know the solution and i understood that, we can do it using complement counting answer would be: ${16 \choose 5} - {9\choose 5} - {7 \choose 5}= 4221$

But can anyone tell me, whats wrong with this ? $\color{red} {{9 \choose 1} \times {5 \choose 1} \times {14 \choose 3}}$ ways. First selected exactly 1 man and 1 woman and then rest.

  • $\begingroup$ Suppose you select A and b initially, and C,d, E later. You could have also selected C,d first and A,b,E later. So there is overcounting. You can't fractionate $\endgroup$ – true blue anil Jan 30 '17 at 9:23
  • $\begingroup$ Yes, I got the mistake now...thnks :) Even if u have provided this as solution, i could have accepted that :) $\endgroup$ – user3699192 Jan 30 '17 at 9:29
  • $\begingroup$ Fine, no sweat ! :) $\endgroup$ – true blue anil Jan 30 '17 at 12:21

It counts multiple times the same team.For example it counts as different team the following 2 teams.

1)First pick John , then Mary and then Jane.

2)First pick John, then Jane and then Mary.

  • $\begingroup$ thnks :) I got it $\endgroup$ – user3699192 Jan 30 '17 at 9:26

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.