Trouble Understanding this Combinatorics Problem Problem:
There are 2 girls and 7 boys in a chess club. A team of four persons must be chosen for a tournament, and there must be at least 1 girl on the team. In how many ways can this be done?
Now,the solution suggested by the book is C(7,2)+2xC(7,3)=91 ,Which I understand,But I don't understand why my answer which is C(2,1)xC(8,3) is wrong. Because after choosing 1 girl,There are no more rules,So we should just choose 3 more people.
 A: Your answer has some double counting. You may choose girl A then when choosing three more people you choose girl $B$ and two boys. You also may choose girl B then when choosing three more people you choose girl $A$ and those same two boys.
Your formula counts that possibility twice but it should count only once.
You could correct your formula by subtracting the double counting, which happens when you choose the two girls and two boys. There are $C(2,7)=21$ ways of doing that, so your corrected formula reads
$$C(1,2)\times C(3,8)-C(2,7)=91$$
which is correct.
By the way, I believe it is more common to write that as
$$C(2,1)\times C(8,3)-C(7,2)=91$$
A: You answer is incorrect because you will have duplicates in your answer.
If you choose a girl A from the girls and then girl B, boy C, and boy D from the rest, that will be the same choosing girl B from the girls and then girl A, boy C, and boy D.  
A: When you see the phrase "at least", it's usually easier to consider the problem the other way round: i.e get the total of all selections minus the selections with only boys, and therefore $$\binom 94-\binom74=91$$I know this doesn't answer your question exactly (this has already been answered by Amy B), but I hope it helps, and it's the kind of thing your book should be suggesting.
