<s>[INST] Children's Fruit Division How many ways can $11$ apples and $9$ pears be divided between 4 children so that each child receives five fruits? (Apples are identical. just like pears).
Solution: $f\left(x,y\right)=\left(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5\right)^4$
$f\left(x,y\right)=\left(\frac{x^6-y^6}{x-y}\right)^4$
$f\left(x,y\right)=\left(x^6-y^6\right)^4{\cdot\left(x-y\right)}^{-4}$
$f\left( x,y \right)={{\left( {{x}^{6}}-{{y}^{6}} \right)}^{4}}\cdot {{x}^{-4}}{{\left( 1-\frac{y}{x} \right)}^{-4}}$
$f\left(x,y\right)=\left(x^{24}-4x^{18}y^6+{6x}^{12}y^{12}-{4x}^6y^{18}+y^{24}\right)\cdot\sum_{k=0}^{\infty}\binom{3+k}{k}x^{-4-k}y^k$
The coefficient of $x^{11}y^9$ in $f\left(x,y\right)=\binom{3+9}{9}-4\binom{3+3}{3}=140.$
I'm right?
 [/INST] Yes that is correct. But a simpler solution exists given each kid must get equal number of fruits. We can first distribute $9$ pears (or $11$ apples) to  four children such that none of them get more than $5$ pears. We then note that as each kid must have five fruits, once we distribute pears, the distributions of apples are fixed.
Unrestricted number of ways to distribute $9$ pears using stars and bars method: $ \displaystyle {{9+4-1} \choose {4-1}} = {12 \choose 3}$
Now we subtract distributions where a kid would have received more than five pears. We choose a kid, assign $6$ pears and then distribute rest $3$ pears among them.
That is given by, $ \displaystyle 4 \cdot {3 + 4 - 1 \choose 4 - 1} = 4 \cdot {6 \choose 3}$
So number of ways to distribute $9$ pears such that no kid receives more than $5$ pears,
$ \displaystyle = {12 \choose 3} - 4 \cdot {6 \choose 3} = 140$
So, number of ways to distribute $11$ apples and $9$ pears such that each kid receives five fruits is also $140$.
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