Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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| − | We can use generating functions, where <math>(x+x^2+...+x^6)</math> is the function for each die. We want to find the coefficient of <math>x^{10}</math> in <math>(x+x^2+...+x^6)^7</math>, which is the coefficient of <math>x^3</math> in <math>\left(\frac{1-x^7}{1-x}\right)^7</math>. This evaluates to <math>\dbinom{-7}{3}=\boxed{\textbf{(E) } 84}</math> | + | We can use generating functions, where <math>(x+x^2+...+x^6)</math> is the function for each die. We want to find the coefficient of <math>x^{10}</math> in <math>(x+x^2+...+x^6)^7</math>, which is the coefficient of <math>x^3</math> in <math>\left(\frac{1-x^7}{1-x}\right)^7</math>. This evaluates to <math>\dbinom{-7}{3} \cdot (-1)^3=\boxed{\textbf{(E) } 84}</math> |
-wannabecharmander | -wannabecharmander | ||
Revision as of 13:39, 3 August 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 ?
Contents
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.
![\[4,1,1,1,1,1,1: 7\]](http://latex.artofproblemsolving.com/f/c/4/fc42b407b8c7d85fa7bcdd7ebf7b4191cf04576e.png)
![\[3,2,1,1,1,1,1: 42\]](http://latex.artofproblemsolving.com/9/9/7/997b2b5435a7b7bfdcc7343d05cb9275f32a7e0a.png)
![\[2,2,2,1,1,1,1: 35.\]](http://latex.artofproblemsolving.com/6/7/1/6714cfcffc2d3e61a7a95c562190f6c0c9025b81.png)
Add up the possibilities: 
.
Thus we have repeated Solution 1 exactly, but with less explanation.
~kevinmathz
Solution 4 (overkill)
We can use generating functions, where 
is the function for each die. We want to find the coefficient of 
in 
, which is the coefficient of 
in 
. This evaluates to 
-wannabecharmander
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. 
