Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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<cmath>2,2,2,1,1,1,1: 35.</cmath> | <cmath>2,2,2,1,1,1,1: 35.</cmath> | ||
| − | Add up the possibilities: <math>35+42+7=\boxed{\textbf{(E) } 84}</math>. | + | Add up the possibilities: <math>35+42+7=\boxed{\textbf{(E) } 84}</math>. |
===Solution 2=== | ===Solution 2=== | ||
| − | Rolling a sum of <math>10</math> 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 dices' 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 | + | Rolling a sum of <math>10</math> 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 dices' 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 <math>\dbinom{9}{3}=\boxed{\textbf{(E) } 84}</math> |
===Solution 3 (overkill)=== | ===Solution 3 (overkill)=== | ||
| − | 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 | + | 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> |
===Solution 4 (Stars and Bars)=== | ===Solution 4 (Stars and Bars)=== | ||
| − | If we let each number take its minimum value of 1, we will get 7 as the minimum sum. So we can do <math>10</math> - <math>7</math> = <math>3</math> to find the number of balls we need to distribute to get three more added to the minimum to get 10, so the problem is asking how many ways can you put <math>3</math> balls into <math>7</math> boxes. From there we get <math>\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{84}</math> | + | If we let each number take its minimum value of 1, we will get 7 as the minimum sum. So we can do <math>10</math> - <math>7</math> = <math>3</math> to find the number of balls we need to distribute to get three more added to the minimum to get 10, so the problem is asking how many ways can you put <math>3</math> balls into <math>7</math> boxes. From there we get <math>\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{\textbf{(E) }84}</math> |
| − | == Solution 5 (Similar to above, using number separation) == | + | |
| + | |||
| + | === Solution 5 (Similar to above, using number separation) === | ||
We can use number separation for this problem. If we set each of the dice value to <math>D\{a, b, c, d, e, f, g, h\}</math>, we can say | We can use number separation for this problem. If we set each of the dice value to <math>D\{a, b, c, d, e, f, g, h\}</math>, we can say | ||
<math>D = 10</math> and each of <math>D</math>'s elements are larger than <math>0</math>. Using the positive number separation formula, which is <math>\dbinom{n-1}{r-1}</math>, we can make the following equations. | <math>D = 10</math> and each of <math>D</math>'s elements are larger than <math>0</math>. Using the positive number separation formula, which is <math>\dbinom{n-1}{r-1}</math>, we can make the following equations. | ||
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\dbinom{9}{3} &= \\ | \dbinom{9}{3} &= \\ | ||
\dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} &= \\ | \dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} &= \\ | ||
| − | 12 \cdot 7 &= \boxed{\textbf{( | + | 12 \cdot 7 &= \boxed{\textbf{(E) }84} \\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Note: We are unable to use non-negative number separations due to the fact that the dice *must* be larger than <math>0</math> or positive. | Note: We are unable to use non-negative number separations due to the fact that the dice *must* be larger than <math>0</math> or positive. | ||
| + | |||
~ Wiselion =) | ~ Wiselion =) | ||
Revision as of 11:59, 6 October 2024
Contents
Problem
When 
fair standard 
-sided dice are thrown, the probability that the sum of the numbers on the top faces is 
can be written as ![\[\frac{n}{6^{7}},\]](http://latex.artofproblemsolving.com/5/a/c/5ac603cdac4dfed9654cab1540ba3f1927ef6202.png)
where 
is a positive integer. What is ?
Solutions
Solution 1
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: 
.
Solution 2
Rolling a sum of 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 dices' 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 
Solution 3 (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 
Solution 4 (Stars and Bars)
If we let each number take its minimum value of 1, we will get 7 as the minimum sum. So we can do - = to find the number of balls we need to distribute to get three more added to the minimum to get 10, so the problem is asking how many ways can you put balls into boxes. From there we get 
Solution 5 (Similar to above, using number separation)
We can use number separation for this problem. If we set each of the dice value to 
, we can say
and each of 
's elements are larger than 
. Using the positive number separation formula, which is 
, we can make the following equations.
Note: We are unable to use non-negative number separations due to the fact that the dice *must* be larger than or positive.
~ Wiselion =)
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solution 1
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=5381
~ pi_is_3.14
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. 
