$\textbf{Q}$ : There are $6$ children each offered a single scoop of any of $3$ flavors of ice cream. In how many ways can each child choose a flavor for their scoop so that some flavor of ice cream is selected by exactly $3$ children?
- I struggled with this for a bit. I'm counting this as $ 3 \cdot C(6,3) \cdot 2 \cdot 2 \cdot 1$ where $3$ is for the number of colors, $C(6,3)$ is choosing $3$ kids from $6$ to have one flavor and $2 \cdot 2 \cdot 1$ comes from the choices that the last $3$ children have.
I'm doing something wrong though as this is not correct. Could someone help point me in the right direction? Thank you.
Best Answer
I think this might be the way @geometry fanatic the mistake you are making is that in the $2.2.1 $ allows the case where two children have the same flavour(other than the flavour assigned to the other $3$). This violates the constraint in the question.So the problem will be solved with the principle of exclusion. So it should be $(3.C(6,3).2.2.2)-6 = 474$. Subtracting $6$ removes the cases where we have two children having the same flavour(of the leftover $2$ flavour). So the other three left children will have the case where all three have same flavour.for eg if we have flavours $f_1 \space f_2 \space and \space f_3$ then $ f_1f_1f_1f_2f_2f_2$ will be a case. I hope it is clear now.