There are 20 indistinguishable children who would like to have one ice cream cone each. There are 6 distinct flavors of ice cream. How many distinct collections of ice cream cones are there where at least two children must order each flavor?
My thought is that we know already there are 12 children choosing the 6 flavors, each from 2 children. So the number of collections relies on the number of the other 8 children so there are $6^{20-12}=1679616$ distinct collections in total. Am i right for the answer?
Best Answer
You forgot that the children are indistinguishable in your final step.
Since the children are indistinguishable, what matters here is how many of them order each flavor. Let $x_i$ be the number of children who order the $i$th flavor, where $1 \leq i \leq 6$. Since a total of $20$ ice cream cones are ordered, $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20 \tag{1}$$ where $x_i \geq 2, 1 \leq i \leq 6$.
Let $x_i' = x_i - 2, 1 \leq i \leq 6$. Then each $x_i'$ is a nonnegative integer. Substituting $x_i' + 2$ for $x_i$, $1 \leq i \leq 6$, in equation 1 yields \begin{align*} x_1' + 2 + x_2' + 2 + x_3' + 2 + x_4' + 2 + x_5' + 2 + x_6' + 2 & = 20\\ x_1' + x_2' + x_3' + x_4' + x_5' + x_6' & = 8 \tag{2} \end{align*} which is an equation in the nonnegative integers.
A particular solution of equation 2 corresponds to the placement of $6 - 1 = 5$ addition signs in a row of eight ones. For instance, $$+ 1 1 + 1 + 1 1 + 1 + 1 1$$ corresponds to the solution $x_1' = 0, x_2' = 2, x_3' = 1, x_4' = 2, x_5' = 1, x_6' = 2$ (and $x_1 = 2, x_2 = 4, x_3 = 3, x_4 = 4, x_5 = 3, x_6 = 4$). The number of such solutions is the number of ways we can place five addition signs in a row of eight ones, which is $$\binom{8 + 6 - 1}{6 - 1} = \binom{13}{5}$$