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A committee of $6$ people will be formed with $7$ men and $6$ women. The oldest of the $7$ men is A and the oldest of the $6$ women is B. It is described that the committee can include at most one of A and B. In how many ways can the committee be chosen?

My attempt:

$13C6- (6C1 \cdot 7C5+6C5 \cdot 7C1) = 1548$

Correct answer is $1386$.

Please help Thanks!

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    $\begingroup$ Welcome to Math SE. Could you share what you have tried so far? $\endgroup$ – A. A. Feb 4 '16 at 10:51
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    $\begingroup$ Do you know how many ways the commitee can be chosen without this constraint? $\endgroup$ – Arthur Feb 4 '16 at 11:08
  • $\begingroup$ I am voting to close since the OP has visited the site since Adolfo posed his query without responding to it. $\endgroup$ – N. F. Taussig Feb 4 '16 at 11:10
  • $\begingroup$ No. of ways=No. of ways with only A included + No. of ways with only B included + No. of ways with neither included. $\endgroup$ – GoodDeeds Feb 4 '16 at 11:14
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    $\begingroup$ Thanks lol hey don't close it I post this before going for dinner =.= wtf, sorry was off to dinner closed the windows righta away,In fact I'm still eating my dinner now =3= 13C6- (6C1*7C5+6C5*7C1)=1548 Correct answer is 1386 $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 11:22
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From all 13C6 solutions, you must subtract the solutions where both A and B are chosen.

If you choose both A and B, there are 11C4 possibilities for the other 4 in the committee.

So 13C6 - 11C4 = 1386.

You calculated the committees where it was not allowed to have exactly one of A and B.

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  • $\begingroup$ @@ thanks a lot, i feel so dumby for not being able to think of this $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 11:34
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Pieter21 has provided you with a valid and efficient solution.

Your attempt was incorrect since you subtracted the sum of the number of ways of selecting five men and one woman and the number of ways of selecting five women and one man. However, it is possible to select a committee of six people containing at most one of A and B that contains more than one person of each sex.

Here is an alternate solution:

A committee that contains at most one of A and B either contains neither A nor B or it contains exactly one of them.

If the committee of six people contains neither A nor B, we must select six of the other eleven available people, which can be done in $$\binom{11}{6}$$ ways.

If the committee contains exactly one of A and B, we must select one of A and B, which can be done in $\binom{2}{1}$ ways, and select five of the other eleven people, which can be done in $\binom{11}{5}$ ways. Hence, the number of committees of six people that contain exactly one of A and B is $$\binom{2}{1}\binom{11}{5}$$

Therefore, the number of committees that can be formed with at most one of A and B is $$\binom{11}{6} + \binom{2}{1}\binom{11}{5}$$

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  • $\begingroup$ Thanks! :) Helpful in helping me understand! $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 15:01

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