Here's an innocent-looking question:

There are 4 girls and 7 boys. How many ways can we select a team of 5, such that at least 3 members are girls?

One way to solve this is to consider the cases when there are exactly 3 girls and exactly 4 girls in the team.

So, when there are exactly 3 girls, # of ways = C(4,3)*C(7,2)= 4*21 = 84

And when there are exactly 4 girls, # of ways= C(4,4)*C(7,1)= 1*7 = 7

And no we just add them up, so total # of ways to select a team with at least 3 girls = 84+7 =91

Simple enough......but

Suppose I take it this way: There need to be at least 3 girls in the team. These three girls can be selected in C(4,3) = 4 ways.

Now, we have 2 vacancies in the team, and 8 people (7 boys + the girl left out) who can fill it. So we can select these 2 team members in C(8,2)= 28 ways.

Thus the total number of ways to select a team should be 4*28 =112

What's going wrong??

I've tried a few more questions like this, and the solutions given by these two methods never match up.... What gives?