Difference between revisions of "2018 AMC 10A Problems/Problem 11"
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− | + | ==Problem== | |
− | <math>\textbf{(A) } | + | When <math>7</math> fair standard <math>6</math>-sided dice are thrown, the probability that the sum of the numbers on the top faces is <math>10</math> can be written as <cmath>\frac{n}{6^{7}},</cmath> where <math>n</math> is a positive integer. What is <math>n</math>? |
+ | |||
+ | <math> | ||
+ | \textbf{(A) }42\qquad | ||
+ | \textbf{(B) }49\qquad | ||
+ | \textbf{(C) }56\qquad | ||
+ | \textbf{(D) }63\qquad | ||
+ | \textbf{(E) }84\qquad | ||
+ | </math> | ||
== Solutions == | == Solutions == | ||
===Solution 1=== | ===Solution 1=== | ||
− | |||
− | + | Add possibilities. There are <math>3</math> ways to sum to <math>10</math>, listed below. | |
− | |||
− | |||
− | |||
− | |||
− | There are <math> | ||
− | + | <cmath>4,1,1,1,1,1,1: 7</cmath> | |
+ | <cmath>3,2,1,1,1,1,1: 42</cmath> | ||
+ | <cmath>2,2,2,1,1,1,1: 35.</cmath> | ||
− | + | Add up the possibilities: <math>35+42+7=\boxed{\textbf{(E) } 84}</math>. | |
===Solution 2=== | ===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 | + | 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> ways. |
− | + | ===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}</math> | ||
− | ===Solution | + | ===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> | |
− | + | ==Video Solution 1== | |
− | + | https://youtu.be/HVn1WV80ZIU | |
− | |||
+ | ~savannahsolver | ||
− | + | == Video Solution by OmegaLearn== | |
+ | https://youtu.be/5UojVH4Cqqs?t=5381 | ||
− | ~ | + | ~ pi_is_3.14 |
== See Also == | == See Also == | ||
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{{AMC10 box|year=2018|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2018|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Probability Problems]] |
Latest revision as of 22:32, 20 November 2022
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 where is a positive integer. What is ?
Solutions
Solution 1
Add possibilities. There are ways to sum to , listed below.
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 ways.
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
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.