# This solution doesn't make sense to me (Probability)

1. Feb 22, 2012

### Xyius

1. The problem statement, all variables and given/known data
A team consisting of three boys and four girls must be formed from a group of nine boys and eight girls. If two of the girls are feuding and refusing to play on the same team, how many possibilities do we have?

2. Relevant equations
Combination formula $\binom{n}{r}=\frac{n!}{(n-r)!r!}$

3. The attempt at a solution
My logic was that there are $\binom{9}{3}$ ways to choose the boys. For the girls I am a bit confused on how I would approach it. There are $\binom{6}{4}$ ways of choosing with only one of the girls who are fighting. But I do not know where to go from here.

The solution says..

$\binom{9}{3}\left[ \binom{6}{4}+2\binom{6}{3} \right]$

Almost got it, but I do not understand the logic behind the second term in the brackets.

2. Feb 22, 2012

### alanlu

The C(6, 4) term is choosing four among the girls who aren't fighting. The C(6, 3) terms chooses three girls from the 6 not fighting, assuming one of the girls fighting is on the team.

3. Feb 22, 2012

### Ray Vickson

If Amy (A) and Brenda (B) are the two feuding girls you can pick girls for the team by (i) exlcuding both A and B; (ii) including A but excluding B; (ii) including B but excluding A.

RGV

4. Feb 22, 2012

### Xyius

Thanks a lot guys! I understand now! :D