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?

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 $$140$$.