An 8-persons committee is to be formed out of a group of 15 women and 12 men. In how many ways can the committee be chosen if there must be at least two men?

Please check my answer

I started the problem by finding all possible committees without gender concern that is


and I am not sure how to find answer directly. I choose to find from the case that violates the condition that is


the answer is


  • $\begingroup$ I have tried to improve your text. In particular a title should be synthetic, not contain the detailed statement of the problem. $\endgroup$ – Jean Marie Oct 8 '17 at 7:07
  • $\begingroup$ Why edit my question ? $\endgroup$ – Lingnoi401 Oct 8 '17 at 7:07
  • $\begingroup$ Um,I have gotten down vote by after someone edit my question without remove full text of my question with and make question is not clear like this Please re-edit my question $\endgroup$ – Lingnoi401 Oct 8 '17 at 7:11
  • 1
    $\begingroup$ The downvote (not mine) has nothing to do with the editing of your question. I am sorry to say that I am pretty certain that the level of english is quite better now than it was at first... I think that the downvote (that I do not approve) is because your solution is far from being exact. (Ctd...) $\endgroup$ – Jean Marie Oct 8 '17 at 7:12
  • $\begingroup$ I mean I had had downvote in a long time ago because it make question was not clear similiar like this question sorry for my bad english skill $\endgroup$ – Lingnoi401 Oct 8 '17 at 7:16
  • $\;\binom{12}{0}\binom{15}{8}$ committees with $0$ men.
  • $\;\binom{12}{1}\binom{15}{7}$ committees with $1$ man.

Hence the number of committees with at least $2$ men is $$ {\small{\binom{27}{8}}}-\left( {\small{\binom{12}{0}}}{\small{\binom{15}{8}}} + {\small{\binom{12}{1}}}{\small{\binom{15}{7}}} \right) $$


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