Basketball Team Combinatorial Question: A basketball team has 5 players, 3 in forward position (which includes a center) and 2 in guard position. In how many ways can we make a team if there are 6 forwards, 4 guards and 2 people who can play forward or guard?
The way I am approaching this question is as follows:
We need to pick 3 people out of 6 for the forward position: $\binom{6}{3}$.
We also need to pick 2 guards out of possible 4: $\binom{4}{2}$.
So far, the answer is: $\binom{6}{3} \times \binom{4}{2}$. (assuming that the 2 people who can play forward or guard statement is disregarded).
If two people can play forward or guard, I am thinking of assuming both play forward, add that to assuming one is forward one is guard, plus assuming two are guards. So:
$\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$
Would this yield the right answer? If not, why?

When you write $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$, you consider the possibility of not using the free positioned player at all, thrice. So, I think you should subtract $2\times \binom{6}{3}\times \binom{4}{2}$ from $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$.
