# 2021 April MIMC 10 Problems/Problem 13

Given that Giant want to put $12$ green identical balls into $3$ different boxes such that each box contains at least two balls, and that no box can contain $7$ or more balls. Find the number of ways that Giant can accomplish this. $\textbf{(A)} ~0 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~19$

## Solution

We can use stars and bars to calculate the total amount of ways without thinking about the condition "no box can contain $7$ or more balls". Giving each box $1$ ball in advance to meet "each box contains at least two balls" requirement since each box will get at least $1$ extra, There are ${8\choose2}$ or $28$ ways to accomplish this. We need to subtract the situations where a box contains $7$ or more balls. There are two situations: $2,2,8$ in each box or $2,3,7$ in each box. In the first situation, since two boxes have the same amount, it will have $\frac{3!}{2}=3$ ways of accomplishing it, and the second situation will have $3!=6$ ways of accomplishing it. Therefore, the total amount of ways that meets the requirement is $28-3-6=\fbox{\textbf{(E)} 19}$.