For part b):
We need to find
$$\tag{1}
\text{number of bowls that have both Chocolate and Vanilla}\over \text{total number of different bowls }
$$
For the denominator of $(1)$: the total number of ways to select five flavors from seven is ${7\choose5}={7!\over 2!\cdot 5!}={7\cdot 6\over 2}=21$. (Generally, the number of ways to select $r$ objects from $n$ distinct objects where the order of selection does not matter is ${n\choose r}={n!\over (n-r)!r!}$).
For the numerator of (1), we need to find the number of ways to choose five flavors with both Vanilla and Chocolate amongst them. Hmm, two of the flavors are already selected: we need only select three flavors from the five flavors that aren't Vanilla or Chocolate.
The number of ways to do this is ${5\choose3}={5!\over2!\cdot 3!}={5\cdot 4\over 2}=10$.
So, computing (1) now, the desired probability is $10\over 21$.
For part c):
We need to compute:
$$\tag{2}
\text{number of bowls that have neither Chocolate nor Vanilla}\over \text{total number of different bowls }
$$
We already computed the denominator of (2) in part b), it is 21.
To find the numerator:
The number of ways of choosing five flavors from the five flavors that aren't Vanilla or Chocolate is ${5\choose5}={1}$.
So, computing (2) now, the desired probability is $1\over21$.
For part d):
We need to compute
$$\tag{3}
\text{number of bowls that have exactly one of Chocolate or Vanilla}\over \text{total number of different bowls }
$$
As before, the denominator in (3) is 21.
To find the number of ways to get a bowl with exactly one of the flavors Vanilla or Chocolate, it proves convenient to break this up into cases:
Case 1): Vanilla only: we have one flavor already (Vanilla); we then need to choose four more flavors that are neither Vanilla nor Chocolate: This is ${5\choose4}={5!\over 1!\cdot 4!}= 5$.
Case 2): Chocolate only: we have one flavor already (Chocolate); we then need to choose four more flavors that are neither Chocolate nor Vanilla: This is ${5\choose4}= 5$.
So,
the number of ways to get a bowl with exactly one of the flavors Vanilla or Chocolate
is obtained by adding the preceeding two quantities: $ 5+ 5=10$.
So, computing (3) now, the desired probability is $10/21$.
Parts e) and f) are somewhat different as they are conditional probabilities.
For part e):
Let
$\ \ \ \ \ \ \ A$ be the event that the bowl has both Chocolate and Vanilla
and let
$\ \ \ \ \ \ \ B$ be the event that at the bowl has at least one of the flavors Chocolate and Vanilla.
We want to compute
$$
P(A|B)={P(A\cap B)\over P(B)}
$$
Now $A\cap B$ is just the event that the bowl has both Vanilla and Chocolate. We found
$P(A\cap B)$ in part b). It turned out to be $P(A\cap B)={10\over 21}$.
To find $P(B)$, we need to find the number of bowls that have at least one of Vanilla or Chocolate. To do this, we break it up into cases:
exactly one of Vanilla or Chocolate: 10 (part d)
both Vanilla and Chocolate: 10 (part b)
So the number of bowls with at least one of Vanilla or Chocolate is $10+10=20$ (note, we could have also computed this by taking the total number of bowls and subtracting the number that had neither Vanilla nor Chocolate: $21-1=20$).
So $P(B)= {10+10\over 21}={20\over21}$.
and
$P(A|B) = { 10/21 \over 20/21}={1\over2}$.
I'll leave part f) for you; but leave a comment if you need help with it.
So we can think your question like this: we can choose a incomplete order, then build the next item to be put on the order, then choose a incomplete order, then build the next item to be put on the order, etc.
So as there are $5$ orders, we can number them $1, 2, 3, 4, 5$. Then we have $3$ $1$'s, $3$ $2$'s $3$ $3$'s $3$ $4$'s and $3$ $5$'s. Then we have $\frac{(3\times5)!}{{3!}^5}=168168000$ ways. That is a lot!
Best Answer
For me, the real challenge is how to reverse-engineer the problem composer's mistaken factor of $(24)$, for the number of ways of choosing the two flavors.
This is my speculative explanation.
There are actually $(5)$ flavors, with the $5$th flavor representing (void). So, you start with an enumeration of $(5)^2$, for the total number of ways of choosing two of the $(5)$ flavors, with replacement.
Here, it is presumed that the order of selection is relevant, in that vanilla on top of chocolate is presumed different than chocolate on top of vanilla.
Then, presumably, the problem composer deducted $(1)$ to get $(25 - 1 = 24)$, reasoning that you can not have both flavors be void. This analysis overlooks the effect of gravity. That is, having the first flavor be void and the second flavor be chocolate results in the same ice cream cone as having the first flavor be chocolate and the second flavor be void.
So, you have to deduct $(4)$ from the problem composer's enumeration, because the problem composer (for example) counted (void + vanilla) as distinct from (vanilla + void).
Here, my suggested enumeration of
$$(5)^2 - 1 - 4 = 20$$
for the number of flavors presumes that (for example) two scoops of vanilla is distinct from one scoop of vanilla.
The alternative way of arriving at the enumeration of $(20)$ would be to start with the assumption that there are exactly $(2)$ scoops, which yields $(4)^2$. Then, you add $(4)$ for the number of ways of having only $(1)$ scoop.