A certain school has $4$ forwards, $4$ guards, $3$ centers and $1$ person who can play as either a forward or a guard. How many different starting lineups can be made?
I came up with 2 answers to this problem. However I don't know which one is right and I can't tell the difference between the two:
Solution 1:
There are two possibilities, X is a forward, in which there is $\binom{5}{2}\binom{4}{2}\binom{3}{1} = 180$ ways of making this starting lineup. X could be a guard as well, which results in the same number, $180$ ways of making the starting lineup. Add together to get $360$ different ways of making this starting lineup.
Solution 2:
There are three possibilities which encompass all possible starting lineups: x is not picked, x is picked as a forward, and x is picked as a guard.
When x is not picked, there is $\binom{4}{2}\binom{4}{2}\binom{3}{1} = 108$ different lineups without x in it.
when x is picked as a forward, you only need to pick one more forward, so there is $\binom{4}{1}\binom{4}{2}\binom{3}{1} = 72$ different lineups with x as forward. the same number will result when you pick x as a guard: $72$. adding $108+72+72$ results in $252$ different lineups.
So the problem is that I can't see the fault in logic in either of my solutions. Which one is the right one?
edit: centers
Best Answer
The problem is that in solution $1$ you are counting the cases in which $X$ does not play twice, once when he doesn't play as a forward and once when he doesn't play as a guard. Therefore the answer should be $360$ minus the number of line-ups in which he doesn't play, this yields $360-108=252$ line-ups.