2018 AMC 10A Problem 11

All $6^7$ outcomes are equally likely, so $n$ is just the number of ordered results (die 1, die 2, ..., die 7) whose total is $10$.

1. Start from the smallest total. Every die shows at least $1$, so the smallest possible total is $7$. To reach $10$ we need $3$ extra points on top of the seven $1$s.
2. How can $3$ extra points be shared out? There are only three patterns:
- one die gets all $3$ extra (it shows $4$, the rest show $1$);
- one die gets $2$ extra and another gets $1$ (they show $3$ and $2$);
- three different dice get $1$ extra each (they show $2$, $2$, $2$).

No die ever goes above $4$, so the limit of $6$ on a die never gets in the way.
3. Count each pattern.
- One die shows $4$: choose which die, $7$ ways.
- One die shows $3$ and another shows $2$: $7$ choices for the $3$, then $6$ for the $2$, so $7 \cdot 6 = 42$ ways.
- Three dice show $2$: choose which three of the seven dice. There are $\binom{7}{3} = \dfrac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ ways.
4. Add. $7 + 42 + 35 = 84$.

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

Taking away the $1$ every die must show turns "dice adding to $10$" into "share $3$ extra points among $7$ dice", which is small enough to list by pattern. In each pattern, count by choosing which dice get the extras; the order matters because the dice are different. Always check whether the top limit ($6$ on a die) can be reached; here it cannot.

Stars and bars. Picture the $3$ extra points as $3$ stars, and split them among $7$ dice using $6$ bars as dividers. Any row of $3$ stars and $6$ bars gives one way to share the points (the stars before the first bar go to die 1, and so on). A row has $9$ spots, and we choose $3$ of them for the stars: $\binom{9}{3} = 84$.