I guess there will be a standard answer to this question somewhere and wouldn't be surprised if this is a duplicate.
However, assuming you've got enough of each sort of candy (which 'big bag of' suggests), each child can have either of the three types of sweets. So, child one has 3 possible candies, child two has 3 possible candies and so on. For the first two children there are $3^2=9$ options this way (here you can still write down the options, if you like) and each successive child adds a factor $3$. Can you see why?
The result then would be $3^8$ ways to distribute candies (which equals $6561$, if I am to believe google). The questoin would be a lot harder if the amount of candies of each type would be restricted.
Your answers to the first two questions are correct.
In how many ways can $15$ indistinguishable candies be distributed to five children if child $C$ and child $D$ receive $7$ candies together?
We must distribute seven candies among the children $C$ and $D$ and eight candies among the children $A$, $B$, and $E$. The number of ways we can distribute the candies to children $C$ and $D$ is eight since $C$ must receive between $0$ and $7$ candies inclusive, with $D$ receiving the rest. The number of ways the remaining eight candies can be distributed to the children $A$, $B$, and $D$ is
$$\binom{8 + 3 - 1}{3 - 1} = \binom{10}{2}$$
as you correctly found. Hence, the number of ways of distributing to the five children if $C$ and $D$ receive exactly seven candies between them is
$$\binom{7 + 2 - 1}{2 - 1}\binom{8 + 3 - 1}{3 - 1} = \binom{8}{1}\binom{10}{2}$$
In how many ways can $15$ indistinguishable candies be distributed to five children if no child receives more than six candies?
Let $x_i$, $1 \leq i \leq 5$, be the number of candies received by the $i$th child. Then we seek the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15 \tag{1}$$
in the nonnegative integers subject to the restrictions that $x_i \leq 6$ for $1 \leq i \leq 5$.
A particular of equation 1 corresponds to the placement of four addition signs in a row of $15$ ones. For instance,
$$1 1 1 + 1 1 1 1 + 1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, $x_3 = 2$, $x_4 = 0$, and $x_5 = 6$. The number of such solutions is the number of ways we can place four addition signs in a row of fifteen ones, which is
$$\binom{15 + 5 - 1}{5 - 1} = \binom{19}{4}$$
since we must choose which four of the nineteen positions required for fifteen ones and four addition signs will be filled with addition signs.
By similar reasoning, the number of solutions of the equation
$$x_1 + x_2 + x_3 + \cdots + x_n = k$$
in the nonnegative integers is
$$\binom{k + n - 1}{n - 1}$$
since we must choose which $n - 1$ of the $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs.
From these, we must subtract those cases in which at least one child receives more than six candies. Observe that at most two children could receive more than six candies since $2 \cdot 7 = 14 < 15 < 21 = 3 \cdot 7$.
Suppose a child receives more than six candies. There are five ways to choose that child. We give that child seven candies. The remaining eight candies can be distributed among the five children in
$$\binom{8 + 5 - 1}{5 - 1} = \binom{12}{4}$$
ways. Hence, there are
$$\binom{5}{1}\binom{12}{4}$$
ways to distribute the candies in such a way that a child receives more than six candies.
However, if we subtract this amount from the total, we will have subtracted too much since we have counted each case in which two children receive more than six candies twice, once for each way of designating one of those children as the child who received more than six candies. We only want to subtract those cases once, so we must add those cases back.
Suppose two children each receive more than six candies. There are $\binom{5}{2}$ ways to select those two children. Give each of them seven candies. That leaves one candy to distribute among the five children, which can be done in five ways. Hence, the number of distributions in which two children receive more than six candies is
$$\binom{5}{2}\binom{1 + 5 - 1}{5 - 1} = \binom{5}{2}\binom{5}{4}$$
By the Inclusion-Exclusion Principle, the number of ways of distributing $15$ indistinguishable candies to five children so that no child receives more than six candies is
$$\binom{19}{4} - \binom{5}{1}\binom{12}{4} + \binom{5}{2}\binom{5}{4}$$
In how many ways can $15$ indistinguishable candies be distributed to five children if each child receives a different amount of candies.
List the ways $15$ can be expressed as a sum of five distinct nonnegative numbers, starting with
$$0 + 1 + 2 + 3 + 9 = 15$$
and ending with
$$1 + 2 + 3 + 4 + 5 = 15$$
For each of these ways (there are not many), there are $5!$ ways to distribute the candies to the children, depending on which child receives which number of candies.
Best Answer
First, "12 Choose 4 ways for each student to receive 3 pieces of candy."
Not at all. If each piece of candy were distinct there would be ${12\choose 3}$ ways for the first child to choose 3 pieces of candy. Then there would be ${9\choose 3}$ ways for the second, etc.
or ${12\choose 3,3,3,3} = \frac {12!}{3!3!3!3!}$
But that is not the problem you have.
I think the easier thing to do is to allocate chocolates, and then assign children to them.
Some child must receive at least 2 chocolates. Putting the chocolate allocations in order from greatest to least, we get
$(3,3,0,0)\\(3,2,1,0)\\(3,1,1,1)\\(2,2,2,0)\\(2,2,1,1)$
are the possible allocations of chocolates.
The gummy bears will be allocated in a complimentary fashion.
Now how to assign children?
How many children receive the same number of chocolates? That makes the denominator for each case.
Case 1. $\frac {4!}{2!2!} = 6$
Case 2. $\frac {4!}{1!1!1!1} = 4! = 24$
Case 3. $\frac {4!}{1!3!} = 4! = 4$
Case 4. $\frac {4!}{3!1!} = 4$
Case 5. $\frac {4!}{2!2!} = 6$
$6+24 +4+ 4 +6 = 44$
Part b) you only have two cases to consider.