Counting how many ways 30 green balls... How many ways are there to distribute 30 green balls to 4 persons if Alice and Eve together get no more than 20 and Lucky gets at least 7?
The answer given to me was $2464 = C(26, 3) − 66 − 46-24$ but I got $C(26, 3) - 6$.
Here is what I did:
Using "stars and bars" (or whatever the actual name for it is), I found the total number of possible combinations. Since 7 balls must immediately go to Lucky, there are 23 balls left to distribute. There are 3 'bars', so the total number of possible combinations is $\binom{26}{3}$.
Because this number accounts for the situations where Alice and Eve have $
\geq 21$ balls, I must subtract those out. (Here is where I think I messed something up). 
When Alice and Eve have 21: This leaves 2 balls left to distribute, so $\binom{3}{2}$
When Alice and Eve have 22: This leaves 1 ball left to distribute, so $\binom{2}{1}$
When Alice and Eve have 23: This leaves no balls left to distribute, so it equals 1.
So $C(26, 3) - 6$.
Where did I go wrong?

Lucky gets $7$ balls right away. Alice and Eve can split their $j\in[0..20]$ balls in $j+1$ ways, then Lucky and Mike can split the remaining $23-j$ balls in $24-j$ ways. The total number of admissible allocations therefore is
$$\sum_{j=0}^{20}(j+1)(24-j)=2464\ .$$
