Children's Fruit Division

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 $140$.