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.
There are two options (assuming all scoops must be distributed):
One kid gets both scoops of chocolate You need to choose the lucky one ($10$) options, then give one vanilla to the rest. You are left with 6 vanilla scoops to give the 10 children. It is equivalent to the number of solutions to $x_1+\ldots+x_{10}=6$ with $x_i\geq 0$, which is ${10+6-1 \choose 10}$.
Two kids get chocolate You need to choose the two (${10 \choose 2}$) then give the 8 kids a scope of vanilla. You are left with 7 vanilla to give the 10 children, so the same as above with different numbers.
Total
$$10 {15 \choose 10} + {10 \choose 2}{16 \choose 10}$$
Best Answer
I think this maybe what you are looking for...
a) Say each of the 6 children were to get one pencil each, then that can be done in only 1 way, since the pencils are all identical, it doesn't matter who gets which one, they are all the same.
or
b) Each of the six children get at-least one pencil.Then this is a stars and bars problem. It's like solving x1 + x2 + x3 + x4 + x5 + x6 = 20, but all xi take positive integer values > 0.Then the answer will be 20-1C6-1 ways.
or
c) It's possible that at-least one kid gets no pencil, which is the same as solving x1 + x2 + x3 + x4 + x5 + x6 = 20, but xi can also be 0 unlike in the previous case. Then this is 20+6-1C6-1 ways.
d) Now to answer, the probability question, like you said this is 1, because if each student gets a different number of pencils, starting with 1 pencil for the first student, 2 pencils for the second, 3 for the the third and so on...then, the least possible number of pencils such that each student gets a different no: of pencils, will be 21, since, 1+2+3+4+5+6 = 21. Therefore, at-least two students have to get the same number of pencils to make the total 20. I guess we can write that as p(atleast two students getting same no of pencils) = 1 - p(everyone getting different no: of pencils) and everyone getting different number of pencils will be => x1 + x2 + x3 + x4 + x5 + x6 = 20, such that, x1 ≠ x2 ≠ x3 ≠ x4 ≠ x5 ≠ x6, and xi >0.