1: This problem is equivalent to putting two identical balls into 16 boxes (with repetition allowed).
$\binom{2+16-1}{2}$
Where 2 is the number of balls, and $16-1$ is the number of separators needed to divided a line segment into 16 smaller segments,each segment represents a box.
Each way of putting the balls (each combination of flavors) will be corresponds to a arrangement of $2$ balls and $16-1$ seperators (where balls and seperators are indistinguishable among themselves).
2.$\binom{3+16-1}{3}$.
A. As shown above, where we assume repetition is allowed.
B. If repetition is not allow,
$\binom{16}{2}$
$\binom{16}{3}$
7 vanilla double-scoops, 14 chocolate double-scoops, and 9 strawberry double-scoops in stock. In addition, the ice cream stand also has combo flavors in stock, there are 10 vanilla/chocolate combo scoops, and 6 vanilla/strawberry combo scoops.
Going with the interpretation that all $\binom{56}{3}$ selections of icecream packages are equally likely (which I still don't like... what icecream salesman doesn't keep their supply organized?) we approach by way of inclusion-exclusion over the events that a particular flavor was missing. Let $X$ be the event that there was no vanilla, $Y$ that there was no chocolate, and $Z$ that there was no strawberry.
You are tasked with finding $\Pr(X^c\cap Y^c\cap Z^c)$ which can be rearranged and expanded into:
$$=1-\Pr(X\cup Y\cup Z) = 1 - \Pr(X)-\Pr(Y)-\Pr(Z)+\Pr(X\cap Y)+\Pr(X\cap Z)+\Pr(Y\cap Z)-\Pr(X\cap Y\cap Z)$$
The probability that no chocolate appeared... well, that is found simply by excluding all cones that have chocolate from our count and selecting from what remains. There are $56$ cones originally, $24$ have chocolate... so $\Pr(Y) = \dfrac{\binom{32}{3}}{\binom{56}{3}}$. $\Pr(X)$ and $\Pr(Z)$ is found similarly.
As for finding $\Pr(X\cap Y)$, here we exclude all cones which have chocolate and/or vanilla in any capacity... leaving only those which are purely strawberry. Here that would be $\Pr(X\cap Y)=\dfrac{\binom{9}{3}}{\binom{56}{3}}$. We can similarly find the others.
For the intersection of all three, well... that is impossible to have avoided all flavors given the supply the ice cream man has with him.
The final probability is then:
$$1 - \left(\dfrac{\binom{23}{3}+\binom{32}{3}+\binom{41}{3}-\binom{9}{3}-\binom{14}{3}-\binom{7}{3}}{\binom{56}{3}}\right)$$
This solution is easily generalized to a larger number of children as well (just replace $3$ at the bottom of each binomial coefficient with the number of kids), something which would have been extremely difficult to do with your original plan.
Best Answer
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 $$10 {15 \choose 10} + {10 \choose 2}{16 \choose 10}$$