Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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3, 2+1, and 1+1+1 | 3, 2+1, and 1+1+1 | ||
| − | There are <math> | + | There are <math>7</math> for Case 1, <math>7*6 = 42</math> for Case 2, and <math>\frac{7*6*5}{3!} = 35</math> for Case 3. |
Therefore, the answer is <math>7+42+35 = \boxed {\textbf{(E) } 84}</math> | Therefore, the answer is <math>7+42+35 = \boxed {\textbf{(E) } 84}</math> | ||
Revision as of 01:54, 14 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 ?
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 
for Case 1, 
for Case 2, and 
for Case 3.
Therefore, the answer is 
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 
ways. (RegularHexagon)
Solution 3
Add possibilities. There are 
ways to sum to 
, listed below.
Add up the possibilities: 
.
Thus we have repeated Solution 1 exactly, but with less explanation.
~kevinmathz
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. 
