Number of ways to distribute 55 red balls and 3 green balls Fifty-five identical red balls and three identical green balls are to be distributed among seven children.
Each child must get at least five balls. In how many ways can this be done?
What I have so far: 
Well, I am guessing that we can figure out the case for the number of ways that just the red balls can be distributed among the children. Since 35 of the balls must already be given to the children, there are 20 balls left over so the number of ways that just the red balls can be distributed is ${20}\choose{7}$. And then you have to figure out the number of ways that the red and green balls combined can be distributed among the children. The three green balls can be distributed among seven children as $3\choose7$. And then you find the number of ways that the left over red balls can be distributed amont these children when some of them have green balls. Is this correct?
 A: HINT: If we were just distributing the $55$ red balls, the answer would be $\binom{20+7-1}{7-1}=\binom{26}6$, not $\binom{20}7$: this is a straightforward stars and bars problem, and the reasoning justifying this calculation is described fairly clearly at the link. (Your realization that we're effectively distributing only $20$ balls is correct, however.)
To solve the actual problem, I would first pretend that all $58$ balls are indistinguishable; the same reasoning would then give you $\binom{23+7-1}{7-1}=\binom{29}6$ possible distributions. Now we have to determine in how many distinguishable ways we can turn $3$ of the $58$ balls red. Each child has more than $3$ balls, so choosing $3$ balls to turn red is basically the same as choosing a way to distribute $3$ indistinguishable balls among $7$ people. Using the stars and bars calculation yields not $\binom37$, but rather ... ?
Now what must you do with that last number and the $\binom{29}6$ in order to get the total number of possibilities?
A: Since $3$ is a small number, we can divide into cases.
Case (i) One child gets all the greens.
Case (ii) One child gets $2$ greens, and another gets $1$.
Case (iii) Three children get $1$ green each,
We analyze Case (i) in fair detail, and leave the other two cases to you.  
The child who gets the greens can be chosen in $\binom{7}{1}$ ways. Give her $2$ reds, and give the others $5$ reds each. So $32$ reds are gone. That leaves $23$ reds, which can be distributed among the $7$ children in $\binom{23+7-1}{7-1}$ ways (Stars and Bars, please see Wikipedia). Thus there are $\binom{7}{1}\binom{29}{6}$ ways to do the distribution so that one child gets all the greens.
A: firstly we should distribute $3$green balls to $7$ children using $\binom{n+r-1} {r-1}$ where n is number of objects and r is number of groups this formula gives us $\binom{9}{3}$ then we should add $32$ red balls to make all groups have 5 balls then we will distribute last 23 balls using the same formula so we should multiply $\binom{9}{3}$ with $\binom{29}{6}$, it is a very big number so I will leave it as it is.$$\binom{9}{3}\times\binom{29}{6}$$ 
