There is a problem I had to solve. The problem:
From among 7 boys and 4 girls we want to choose a 6-member volleyball team that has at least 2 girls. In how many ways this can be done?
So I thought, that first I need to find in how many ways I can choose 2 girls. This is ${4 \choose 2}$. So there are 9 people left. And as I already satisfy "at least 2 girls" condition, I can find the number of ways to complete the team, i.e. for each selected 2 girls I can choose left 4 members from 9 people in ${9 \choose 4}$ ways. Then by addition rule there are ${4 \choose 2} {9 \choose 4}$ ways to make a team. But this is not correct answer as I found then. The book I read says there are ${4 \choose 2}{7 \choose 4} + {4 \choose 3}{7 \choose 3} + {4 \choose 4}{7 \choose 2}$ ways.
I understand the last solution, even if I suddenly figured out this solution, I would think that I fully understand the problem and found the solution, as I did with my solution above. I just can't understand why my solution gives different result. Because after knowing the right answer, I expected both of them to give the same result - like different approaches to the problem. But it turns out that I am wrong. And I can't understand what is wrong with my reasoning. Can someone help me to understand what is wrong with the first solution and what is wrong in my reasoning?
Suppose the girls are $A,B,C,D$ and the boys are $T,U,V,W,X,Y,Z$.
And you chose $A$ and $B$ and then choose $T,U,C,V$. You count that once.
And then suppose you chose $A$ and $C$ and then chose $T,U,B, V$. You count that once.
So counted the team $A,B,C,T,U,V$ twice.
Even worse, the team, $A,B,C,D,T,U$ was counted when you picked $A,B$ and then $C,D,T,U$ and when you picked $A,C$ and then $B,D,T,U$ and when you picked $A,D$ then $B,C,T,U$ and when you picked $B,C$ and then $A,D, T,U$, when you pick $B,D$ and then $A,C, T,U$, and when you picked $C,D$ and then $A,B,T,U$. So you counted the team $A,B,C,D,T,U$ six times!