You have overcounted the number of ways to distribute the toys. Suppose the toys are $T_1,T_2,T_3,T_4$. Then one of the toy distributions you counted is where you pick out $T_1,T_2,T_3$ and distribute them to person 1, person 2, and person 3, in that order, and give $T_4$ to person 1. But you've also counted, as a separate way of distributing the toys, the distribution where you pick out $T_2,T_3,T_4$ and give $T_2$ to person 2, $T_4$ to person 3, and $T_4$ to person 1, and then give $T_1$ to person 1. This is, however, the same toy distribution as the previous one.
You can count the number of ways to distribute the toys as follows. Exactly one person must receive two toys, and there are $3$ possibilities for who that person is. There are ${4 \choose 2}=6$ choices for which toys that person receives. There are two remaining toys and two people who must each receive one of them, so there are $2$ ways to distribute the last two toys. This gives $3 \cdot 6 \cdot 2 = 36$ ways to distribute the toys.
Now you can multiply as you did before to get $6 \cdot 36=216$ ways to distribute the marbles and toys.
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
If the marbles are all distinct, then your analysis for the first problem is right. (If the marbles are identical, then the number of ways is $\binom{8}{4}$.)
Let's look at the second problem. We have $11$ identical apples. Line them up like this: $$A\quad A \quad A\quad A \quad A \quad A \quad A \quad A \quad A\quad A \quad A,$$ with a gap between consecutive apples. There are $10$ interapple gaps. We will do our distribution as follows. We will choose $5$ of these gaps to put a separator (say a grape) into. Child $1$ will get all the apples from the leftmost one to the first grape. Child $2$ gets all the apples from the first grape to the second grape. Child $3$ will get all the apples from the second grape to the third, and so on. Finally, Child $6$ gets all the apples from the fifth grape to the right end.
There are exactly as many ways to distribute the apples between the kids as there are to place $5$ grapes as described above. For if we know how many apples each of the kids gets, we know where to place the grapes. Also, as we saw above, every grape placement determines what each child gets. It follows that the number of ways to distribute the apples is $\binom{10}{5}$.
Remark: For completeness, we deal with the marble problem on the assumption marbles are identical. The idea is similar to the apple one, but with a twist. We want to distribute $4$ marbles between $5$ pockets (children). What is different is that we are not asking that each child get a marble.
Imagine doing the distribution as follows. We will distribute $9$ marbles among $5$ children, so that each child gets at least one marble. Then we take away a marble from each child. We will end up with $4$ marbles distrbuted among $5$ children.
Use the separator idea. There are $\binom{8}{4}$ to distribute $9$ identical marbles between $5$ children, with each child getting at least one marble. So there are $\binom{8}{4}$ ways to distribute $4$ marbles among $5$ children.
Actually, with pockets and identical marbles, the numbers are small enough that we could count more crudely. We could put all the marbles in one pocket. There are $\binom{5}{1}$ ways to do this. We could put $4$ marbles in one pocket and $1$ in another. The lucky pocket can be chosen in $\binom{5}{1}$ ways, and for each of these, the pocket that will have one marble can be chosen in $\binom{4}{1}$ ways, for a total of $\binom{5}{1}\binom{4}{1}$. Continue.