How many ways are there to distribute 30 green balls to 4 persons? 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 is: 2464 but I'm not sure how to get it?
 A: Lets say that this represents some distribution of balls to 4 people.
****|**|*****|****
I only have 15 balls in this picture, but you hopefully get the point
Now think of this not as balls and people but as stars and bars.
there are 18 objects in my picture (or 33 in your problem) 3 of which are bars.
The total number of ways to put $n$ objects in $m$ bins, then is ${n-1\choose m-1}$
Now we have some addional criteria.  Alice and eve get no more than 20.  And lucky gets at least 7.
Lets fist give 7 to Lucky.  And then we can remove them from consideration... that leaves 23 balls to distribute.
${26\choose 3}=2600$
And then we need to remove the cases when where Alice and Eve get more than 20.  
Suppose we give Alice and Eve 20.  Lucky gets at least 7, that is 27 of 30 balls accounted for.  There are cases.  21 balls to A+E, 22 balls to A+E, 23 balls to A+E.
${22\choose 1}\cdot{3\choose 1} + {23\choose 1}\cdot{2\choose 1} +  {24\choose 1}\cdot{1\choose 1} = 136$
$2600-136=2464$  
A: If you are familiar with stars and bars, you can do it like this, although I'm sure there is a more efficient way.
First, just give Lucky $7$ balls, leaving $23$ to distribute to the $4$ people. There are $2600$ ways to do that.
But since Alice and Eve can only have a max of $20$, that means if the other two people get $2$ or fewer, that distribution is illegal.
There are six ways that can happen. The balls can be distributed to them 2-0, 1-1, 0-2, 1-0, 0-1, 0-0.  Those cases leave, respectively 21, 21, 21, 22, 22, 23 balls for Alice and Eve.  And then respectively, those balls for Alice and Eve can be distributed 22, 22, 22, 23, 23, 24 ways.
Subtract those illegal distributions from $2600$ to get $2464$.
A: Take out seven to give to Lucky.  Now you have $23$ to distribute with no condition on Lucky, but the Adam and Eve constraint is live.  
Note:  if A & E get $n$ between then, then we are distributing $n$ to them ($n+1$ ways to do it) and $23-n$ to L & X, $24-n$ ways to do it.  Hence the answer is $$\sum_{n=0}^{20} (n+1)\times(24-n) = \fbox {2464}$$
A: As lulu said, this is the same as distributing $23$ balls among $4$ people with the similar conditions, since you can just give Lucky $7$ before distributing. If you're familiar with stars and bars problems (assuming that people can get zero balls), there are $26\choose 3$ ways, or $2600$ to distribute $23$ balls among $4$ people. 
However, we want to make sure that Alice and Eve don't get more than $20$ total. Then count how many ways that they can get more than $20$ and then subtract that. As a hint, if they get more than $20$ balls, how many can they have? There should be $3$ cases. For each case, count the number of ways to distribute those balls among Alice and Eve and multiply that by the number of ways to distribute the remaining balls among Lucky and our unnamed friend. Then subtract that from $2600$ for each case, and you'll get your answer.
