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?
 A: I would do this carefully, the long way, and then see whether I have missed any shortcuts. A procedure that does not involve crossing the fingers has high priority!
You counted correctly the number $\binom{6}{3}\binom{4}{2}$ of teams made up of specialists. So there are $120$ such teams. Now we count separately the teams that contain $1$ versatile player, and $2$ versatile players.
If we are going to have $1$ versatile player, she can be chosen in $\binom{2}{1}$ ways. She can replace a forward, leaving $\binom{6}{2}\binom{4}{2}$ choices for the rest of the team, or she can replace a guard, leaving $\binom{6}{3}\binom{4}{1}$ choices for the rest of the team. Thus there are 
$$\binom{2}{1}\left(\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{4}{1}\right)$$
teams with exactly $1$ versatile player.  So there are $340$ such teams. 
If we are going to use $2$ versatile players, they can be both replace forwards, leaving $\binom{6}{1}\binom{4}{2}$ choices, or both replace guards, leaving $\binom{6}{3}$ choices, or do one of each, leaving $\binom{6}{2}\binom{4}{1}$ choices, for a total of 
$$\binom{6}{1}\binom{4}{2}+\binom{6}{3}+\binom{6}{2}\binom{4}{1}$$
teams with exactly $2$ versatile players. So there are $96$ such teams.
Finally, add up. 
Remark: I interpreted team to mean a set of $5$ people. If by team we mean a set of $5$, together with a specification of what positions (forward or guard) they are playing, the answer would be different, since one versatile player in each position would have to be counted twice, once for X playing forward and Y playing guard, and once for the reverse. And one can complicate things further.
A: Note: For this answer, I interpret "team" to mean a set of five players together with their position. Thus, in my counts, it makes a difference whether a versatile player is playing guard or forward.
If you just lump your flexible players into each category, you have 8 forwards and 6 guards. So, a rough estimate for the number of teams is $\binom{8}{3}\binom{6}{2}$. The problem with this is some configurations feature the same person in two positions, which is unacceptable. If we subtract those, we'll have the correct count.
Let $A$ be the set of teams featuring the first flexible player twice and let $B$ be the set of teams featuring the second flexible player twice. We want to find $|A \cup B|$, which is $|A| + |B| - |A \cap B|$ by inclusion-exclusion.
Now, $|A| = \binom{7}{2}\binom{5}{1}$, since we must fill out the team under the assumption that the first player is already playing as both guard and forward. Similarly, for $|B|$. Finally, $|A \cap B| = \binom{6}{1}$, since both the flexible players are playing forward and guard.
Gathering everything up, we have the number of teams being
$$
\begin{align*}
\binom{8}{3}\binom{6}{2} - |A \cup B| &= \binom{8}{3}\binom{6}{2} - \left(2\binom{7}{2}\binom{5}{1} - \binom{6}{1}\right)\\
&= 636.
\end{align*}
$$
A: 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 $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2}$.
