2018 AMC 10A Problem 11

There are $6^7$ equally likely ordered outcomes, so $n$ is simply the number of ordered $7$-tuples $(d_1, \dots, d_7)$ with each $1 \le d_i \le 6$ and $d_1 + \cdots + d_7 = 10$.

Each die shows at least $1$, so set $e_i = d_i - 1 \ge 0$. The condition becomes
$$e_1 + e_2 + \cdots + e_7 = 10 - 7 = 3.$$

The upper bound is automatic: no $e_i$ can exceed $3$, so no die exceeds $4$, well under $6$. We are simply distributing $3$ identical pips among $7$ dice.

By stars and bars, $3$ stars and $6$ bars can be arranged in
$$\binom{3 + 6}{6} = \binom{9}{3} = 84$$
ways.

The answer is $\boxed{\textbf{(E)}\ 84}$.

A sum of dice is a distribution problem once the mandatory $1$ on each die is removed. The shift $e_i = d_i - 1$ converts "at least 1" into "at least 0," which is exactly the setting stars and bars handles. Always check whether the upper limit $6$ can actually be reached; when the leftover total is smaller than $6$, it cannot, and no inclusion-exclusion correction is needed.

Casework on the pattern of extra pips: three dice each show $2$ ($\binom{7}{3} = 35$ ways), one die shows $3$ and another $2$ ($7 \cdot 6 = 42$), or one die shows $4$ ($7$). Total $35 + 42 + 7 = 84$.

• Forgetting that dice cannot show 0 and counting solutions to a sum of 10 over 7 nonnegative variables, or worrying about the upper bound 6 when it cannot bind.
• Using $\binom{9}{2}$ or $\binom{10}{3}$ from a misremembered stars-and-bars formula; the count is $\binom{\text{stars} + \text{bars}}{\text{bars}}$ with $7 - 1 = 6$ bars.