For the question
The $11$ men and $10$ women of the Math Club (including Jack and Jill) need to form a committee consisting of $4$ people. In how many ways can this be done so that there are the same number of women as men, but Jack and Jill are not both on the committee.
I know I can do this question as a complement in that I select total number of combinations possible when there are no restrictions other than equal number of men and women ${11 \choose2} \cdot {10\choose 2}$ and subtract from it the combinations possible when Jack and Jill are on the committee ${9\choose1} \cdot {10\choose1} $ for a total of $2,385$ combination
But I want to try to do it the other way where I list out the combinations, but I am not getting the same answer. I am over counting. Can someone point out where I am going wrong?
Edit I have updated for the earlier error based on @vladz comments
- Combinations when Jack is on the committee but not Jill $$ 10 \cdot {9\choose2} = 360$$
- Combinations when Jill is on the committee but not Jack $$ {10 \choose 2} \cdot 9 = 405$$
- Combinations when neither Jill nor Jack is on the committee $$ {10 \choose 2} \cdot {9 \choose 2} = 1620$$
For which the total is $360 + 405 + 1620 = 2,385$