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.
It turns out there is a simple rephrasing of the problem that makes it easy to count. Rather than write down a list of numbers like (3,5,2) that add to 10, you can write down 10 objects, and place dividers that separate the 10 objects into groups:
* * *|* * * * *|* *
Exercise: convince yourself of the following:
- given any list of numbers that add to 10, you can draw a corresponding diagram like the above
- given any diagram like the above, you can find a corresponding list of numbers that add to 10
- If you have a list of numbers, convert it to a diagram then back to a list, you get your original list
- If you have a diagram, convert it to a list, then back to a diagram, you get the original diagram
There is an easy way to count how many diagrams can be drawn: there are 9 places where we can put a divider, and for each place, we have a choice to put a divider there or not. Therefore, there are $2^9 = 512$ possible choices.
This sort of diagram is called "stars and bars", and there are a number of kinds of problems that can be solved by finding a way to convert back and forth between the problem you want to solve and diagrams like the one above.
If you want to count specifically the number of lists that have a given number of entries, we can do something similar. e.g. if we want there to be exactly three positions (each with at least one person), then we want to count the number of ways to insert two dividers into a list of 10 objects.
Again, convince yourself that you can translate back and forth between lists and diagrams. It can be tricky to get these arguments exactly right sometimes.
If we want there to be $k$ positions, then we want to place $k-1$ dividers into the 9 possible places, and thus there are $\binom{9}{k-1}$ choices.
If you haven't seen binomial coefficients (a.k.a. combinations) before, this is equal to
$$ \binom{n}{r} = \frac{n!}{r! (n-r)!} $$
where $s!$ means the factorial of $s$: that is,
$$ s! = 1 \cdot 2 \cdot \ldots \cdot s $$
(and $0! = 1$)
Best Answer
Consider cases depending on the number of players who can play either position are used.
Suppose exactly $k$ players who can play either position are used, where $0 \leq k \leq 4$. If $d$ of these $k$ players are selected to play defense, where $0 \leq d \leq 2$, then the remaining $k - d = f$ of these players must play forward, where $0 \leq f \leq 3$. If $d$ of the players who could play two positions play defense, then the remaining $2 - d$ defensive players must be selected from the $7$ players who only play defense. If $f$ of the players who could play two positions play forward, then the remaining $3 - f$ forwards must be selected from $10$ players who only play forward. The goalie must be selected from one of the three players who only play goalie. Hence, if $k$ players who could play either defense or forward are selected to play on the team and $d$ of these $k$ players are selected to play defense, then the number of ways to form such a team is $$\binom{3}{1}\binom{4}{k}\binom{k}{d}\binom{7}{2 - d}\binom{10}{3 - f}$$ I will leave it to you to finish the calculations.