You can give any combo of $0-5$ apples, $0-10$ mangoes to person $A$
Thus $6\times11 = 66$ ways (the balance needed to make $15$ will be oranges)
There is a better approach. Let $x_k$ be the number of apples received by the $k$th child. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10$$
is an equation in the positive integers. A particular solution corresponds to the placement of five addition signs in the nine spaces between successive ones in a row of ten ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance,
$$1 1 + 1 + 1 1 1 + 1 + 1 + 1 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 1$, $x_3 = 3$, $x_4 = x_5 = 1$, $x_6 = 2$. The number of such solutions is the number of ways we can select five of the nine spaces in which to place an addition sign, which is
$$\binom{9}{5}$$
Addendum: Using Barry Cipra's observation, we can confirm this result by using your method.
One child receives five apples and the other five children each receive one apple: There are $6$ ways to select the child who receives five apples.
One child receives four apples, another child receives two apples, and each of the other four children each receive one apple: There are six ways to choose the child who receives four apples and five ways to choose the child who receives two apples. Hence, there are $6 \cdot 5 = 30$ such distributions.
Two children each receive three apples and the other four children each receive one apple: There are $$\binom{6}{2} = 15$$ ways to select the two children who each receive two apples.
One child receives three apples, two children each receive two apples, and the other three children each receive one apple: There are six ways to choose the child who receives three apples and $\binom{5}{2}$ ways to choose which two of the other five children each receive two apples. Hence, there are $$\binom{6}{1}\binom{5}{2} = 6 \cdot 10 = 60$$ such distributions.
Four children each receive two apples and the other two children receive one apple: There are
$$\binom{6}{4} = 15$$
ways to select which two children will receive two apples.
Observe that
$$6 + 30 + 15 + 60 + 15 = \binom{9}{5}$$
Best Answer
Yes, you are correct. Distributing $n$ alike fruits among $k$ children: we count the number of nonnegative integer solutions of $$x_1+x_2+\dots +x_k=n\implies \binom{n+k-1}{k-1}.$$ So, in your case, by the multiplication theorem, the total number of ways is $$\binom{4}{2}\cdot \binom{6}{2}=6\cdot 15=90.$$