2018 AMC 10A Problem 11

We can use stars and bars to find $n.$ It is the same as finding the number of ways to put $10$ balls into $7$ boxes, where each box has at least one ball. The formula for such a scenario is $$\binom{n - 1}{k - 1},$$ where $n$ is the number of balls and $k$ is the number of boxes. No die can exceed $4$ in a sum of $10$ from seven positive rolls, so the upper bound of $6$ creates no additional restriction. The desired answer is therefore $$\binom{9}{6} = \binom{9}{3} = 84.$$ Thus, E is the correct answer.