Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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There are <math>\dbinom {7}{1}</math> for Case 1, <math>7*6 = 42</math> for Case 2, and <math>\dbinom {7}{3}</math> for Case 3. | There are <math>\dbinom {7}{1}</math> for Case 1, <math>7*6 = 42</math> for Case 2, and <math>\dbinom {7}{3}</math> for Case 3. | ||
| − | Therefore, the answer is <math>7+42+35 = \boxed {E) 84}</math> | + | Therefore, the answer is <math>7+42+35 = \boxed {(E) 84}</math> |
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| + | Solution by PancakeMonster2004 | ||
== See Also == | == See Also == | ||
Revision as of 15:47, 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
