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 $n$ ?

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

Solutions

Solution 1

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 to 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 $7$ for Case 1, $7\cdot 6 = 42$ for Case 2, and $\frac{7\cdot 6\cdot 5}{3!} = 35$ for Case 3.

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

Solution by PancakeMonster2004

Solution 2

Rolling a sum of 10 with 7 dice can be represented with stars and bars, with 10 stars and 6 bars. Each star represents one of the dots on the die's faces and the bars represent separation between different dice. However, we must note that each die must have at least one dot on a face, so there must already be 7 stars predetermined. We are left with 3 stars and 6 bars, which we can rearrange in $\dbinom{9}{3}=\boxed{\textbf{(E) } 84}$ ways. (RegularHexagon)

Solution 3

Add possibilities. There are $3$ ways to sum to $10$ , listed below.

$4,1,1,1,1,1,1: 7$ $3,2,1,1,1,1,1: 42$ $2,2,2,1,1,1,1: 35.$

Add up the possibilities: $35+42+7=\boxed{\textbf{(E) } 84}$ .

Thus we have repeated Solution 1 exactly, but with less explanation.

~kevinmathz

Solution 4 (overkill)

We can use generating functions, where $(x+x^2+...+x^6)$ is the function for each die. We want to find the coefficient of $x^{10}$ in $(x+x^2+...+x^6)^7$ , which is the coefficient of $x^3$ in $\left(\frac{1-x^7}{1-x}\right)^7$ . This evaluates to $\dbinom{-7}{3} \cdot (-1)^3=\boxed{\textbf{(E) } 84}$

-wannabecharmander

HOW DO YOU DO THIS

Solution 5 (Stars and Bars)

Since every number has to be positive we can subtract $7$ from the $10$ we need to get, so the problem is asking how many ways can you put $3$ Balls into $7$ boxes. From there it is $\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{84}$ (Zachdog1)

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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