2018 AMC 10A Problems/Problem 11

When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as $\frac{n}{6^7},$ where $n$ is a positive integer. What is $n$ ?

$\textbf{(A) } 42 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 84$

Solution

The minimum number that can be shown on the face of a die is 1, so the least possible sum of the top faces of the 7 dies is 7.

In order for the sum to be exactly 10, 1-3 dices' number on the top face must be increased by a total of 3.

There are 3 ways to do so: 3, 2+1, and 1+1+1

There are $\dbinom {7}{1}$ for Case 1, $7*6 = 42$ for Case 2, and $\dbinom {7}{3}$ for Case 3.

Therefore, the answer is $7+42+35 = \boxed {(E) 84}$

Solution by PancakeMonster2004

2018 AMC 10A (Problems • Answer Key • Resources)
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2018 AMC 10A Problems/Problem 11

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