Combinations Question? Fun fun?! There are 10 players on the basketball team, 12 players on the volleyball team, and 15 members of the track team. Of those players, 2 athletes are on all three teams. How many committees of 4 players can be formed if there must be at least 1 member from each team represented? Order doesn't matter. 
attempt:
I got 33 because I subtracted 2 from each team, and then added it to the total (31+2 = 33)
33C4 - 23C4 - 21C4 - 18C4 - 8C4(all bball) - 10C4(all vball) - 13C4(all tennis)
= 22,025
But apparently the answer is 24,015
Am i doing something wrong?

There are $2$ versatiles. We will assume that apart from that, there is no overlap between teams.  
We interpret the problem as meaning that if one or two of the versatiles is on the committee, then the condition that every team is represented is fulfilled.
There are $31$ non-versatiles.
Committees with $2$ versatiles: $\binom{31}{2}$.
Committees with $1$ versatile: $(2)\binom{31}{3}$.
Finally, we must count the committees with no versatiles. There are respectively $8$, $10$, and $13$ on the various teams. With no restrictions that gives $\binom{31}{4}$ committees. From this we must subtract the number of forbidden committees. The number of forbidden committees is
$$\binom{18}{4}+\binom{23}{4}+\binom{21}{4}-\binom{8}{4}-\binom{10}{4}-\binom{13}{4}.$$
(Be careful about the minus signs!)
Remark: We computed using the above expressions (well, the real work was done by a battery powered assistant). The numerical result is $24015$. So the interpretation mentioned at the beginning of the answer was probably the intended one. 
