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Question, with 30 members in a club: 18 are men & 12 are women.
A committee is formed with 6 members, but a requirement that there be at least as many as women as men. How many different committees can be formed?

Kind of confused with the word of the questioning, but I have attempted that I believe they are asking 12 women at least. So - 12C6 = 2310 committees can be formed? Not sure if its correct.

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First off, this is not a programming question but a mathematical problem.

Second, I'm not sure if I understand the question correctly but here's my interpretation: Because you are required to have at least as many women as men on a committee and since there are 6 members in each committee it follows that the minimum number of women in a committee would have to be 3. So there are only 4 possible types of committees:

  • 3 male : 3 female
  • 2 male : 4 female
  • 1 male : 5 female
  • 0 male : 6 female (as pointed out by Philip Sheard)

Then, you could use the following logic:

For committees formed of 3 male and 3 female, the number of possible committees would be the number of possible combinations of elements from the two sets:

  • The set of all possible combinations of 3 males from the total 18 =C(18,3)
  • The set of all possible combinations of 3 females from the total 12 =C(12,3)

Let's call CM = C(18,3) the number of different combinations of 3 men can be formed from the 18 male members; And let's call CF = C(12,3) the number of different combinations of 3 women can be formed from the 12 female members.

The result for a 3:3 type committee would be CF x CM

CF x CM = C(18,3) * C(12,3)
CF x CM = (18! * 12!) / (3! * 15! * 3! * 9!)

Then, you could use the same logic to compute the number of committees for 2:4 and then for 1:5, and finally for 0:6.

Summing them all should give you the final result.

于 2013-10-11T22:41:27.397 回答
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In your notation, the answer is 12C3 * 27C3. Choose 3 women at random, then choose any 3 other people.

于 2013-10-11T22:45:09.973 回答