Counting ways to distribute 7 apples and 5 pears between 4 children with restriction I was given the following problem:

In how many ways $7$ apples and $5$ pears can be distributed between four kids if each child gets three fruits?

My counting is not the strongest, here is how I approached the problem: first I will distribute the pears then fill the missing spots with apples. We can use stars and bars to count the distribution of pears: $\binom{5+4-1}{4-1}=\binom 8 3=56$. We need to subtract cases when there are more than $3$ pears: $4$ (one basket with $5$ pears) and $\frac{4!}{2!}$ ($4$ pears in one basket and one in some other basket). Thus, the total number is $56-4-12=40$. Is this correct? Is there a better way to explain this problem to students?

I'll do the apples rather than the pears to illustrate the simplification I have in mind.  There are $\binom {10}{3}=120$ ways to distribute $7$ apples among $4$ kids without restriction.  If no kid can have $4$ or more apples, then after assigning $4$ of the apples to one of the kids, there are $\binom 63=20$ ways to distribute the remaining $3$ apples among the $4$ kids, so there are $4 \cdot 20=80$ disallowed combinations, meaning $40$ allowed combinations.
