Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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==Solution 2== | ==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>. | ||
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| + | <math>\dbinom {9}{6}</math> | ||
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| + | Solution by Crosscut | ||
== See Also == | == See Also == | ||
Revision as of 16:36, 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 ![\[\frac{n}{6^7},\]](http://latex.artofproblemsolving.com/b/2/0/b20698c3d950be8381a13060106e98af36026d7b.png)
where 
is a positive integer. What is ?
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 
for Case 1, 
for Case 2, and 
for Case 3.
Therefore, the answer is 
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 
numbers and get 
.
Solution by Crosscut
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
