Difference between revisions of "2018 AMC 10A Problems/Problem 11"

(Solution 2)
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Solution by PancakeMonster2004
 
Solution by PancakeMonster2004
 
==Solution 2==
 
The problem can be converted to a ball-and-urn problem. The objective is to find the number of ways there are to sum up <math>7</math> numbers and get <math>10</math>.
 
 
<math>\dbinom {9}{6}</math>
 
 
Solution by Crosscut
 
  
 
== See Also ==
 
== See Also ==

Revision as of 17:41, 8 February 2018

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

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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