1. ## Combinations

solved

2. I'm assuming in all of these that when you say 'starting lineup', you are referring to the final team and not just the effect the problem has on it.

A) $\displaystyle \binom{2}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1} = 2*6*6*5 = 360$

One center out of 2 possible, 2 guards out of 4 possible, then you have 1 power forward out of 6 possible, and 1 small forward out of the remaining 5 forwards.

B) $\displaystyle \binom{1}{1}\binom{5}{2}\binom{6}{2} = 1*10*15 = 150$

You only have 1 center, then you have 5 possible forwards, of which you need 2. What's left is 3 guards and 3 forwards, of which you need 2.

C) The wording here is rather poor, I take it to mean 'is the same result achieved in problem B if the guards are free to change positions rather than the forwards?'. As it does not specify if your players are sick or not, I may be doing the wrong problem.

$\displaystyle \binom{1}{1}\binom{3}{2}\binom{6}{2} = 1*3*15 = 45$

You have your one center, then 3 choices to pick from for 2 guards. You have 1 guard remaining to toss into the forward position, so 6 players to pick 2 from. Therefor: No, the same result is not achieved.