Question 733245
A committee consisting of at least three members is to be formed from a group
of six boys and six girls such that it always has a boy and a girl. Find the
number of ways to form such committee.
<pre>

First we count all committees of any size without regard to size or sex.
(This will even include a committee of size 0.)  Then we'll subtract the 
number of unacceptable committees.

In choosing a committee without regard to size or sex, we 
tell each of the 12 people one of two things:

1.  YES, you will serve on the committee
2.  NO, you will NOT serve on the committee.

If there were no restrictions with regard to size or sex the answer 
would be 2<sup>12</sup>.

Now we subtract the unacceptable committees.

We may not tell all the girls NO, for that would leave an all-boy committee.
So that is the number of ways we could tell all the girls NO, and tell the 6
boys YES or NO, which would be  2<sup>6</sup>.  

We also may not tell all the boys NO, for that would leave an all-girl
committee. So that is the number of ways we could tell all the boys NO, and
tell the 6 girls YES or NO, which would also be  2<sup>6</sup>.

That would be 2·2<sup>6</sup> or 2<sup>7</sup> to subtract. However both those
have 1 case in common, the case where all 12 people are told NO, so we must
avoid subtracting that case twice, so we can only subtract 1 less than that or
2<sup>7</sup> - 1  

So without regard to the committee size, there are

2<sup>12</sup> - (2<sup>7</sup> - 1) or 2<sup>12</sup> - 2<sup>7</sup> + 1 committees.

Now we only have to subtract the number of committees of size 2 consisting
of 1 boy and 1 girl.  There are 6 ways to choose the girl and 6 ways to
choose the boy.  That's 6·6 or 36 committees of 2 to subtract.

[Note: We have already subtracted the committees of sizes 0 and 1. The
committee of size 0 was the one we had to avoid counting twice above, and a
committee of 1 cannot contain both sexes.]

Final answer:  2<sup>12</sup> - 2<sup>7</sup> + 1 - 36 = 4096-128+1-36 = 3933

Edwin</pre>