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 == | ||
Line 12: | Line 20: | ||
3, 2+1, and 1+1+1 | 3, 2+1, and 1+1+1 | ||
− | There are <math>7</math> for Case 1, <math>7 | + | There are <math>7</math> for Case 1, <math>7\cdot 6 = 42</math> for Case 2, and <math>\frac{7\cdot 6\cdot 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> | ||
− | |||
− | |||
===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 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 <math>\dbinom{9}{3}=\boxed{\textbf{(E) } 84}</math> ways. | + | 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 <math>\dbinom{9}{3}=\boxed{\textbf{(E) } 84}</math> ways. |
===Solution 3=== | ===Solution 3=== | ||
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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 | + | Add up the possibilities: <math>35+42+7=\boxed{\textbf{(E) } 84}</math>. |
Thus we have repeated Solution 1 exactly, but with less explanation. | Thus we have repeated Solution 1 exactly, but with less explanation. | ||
− | ~ | + | ===Solution 4 (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 5 (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> | ||
+ | |||
+ | ===Solution 6 (Solution 5 but more clearer and compact) === | ||
+ | Assume each die has value 1. Then we have <math>10-(1 \cdot 7)=3</math> left. This is to be split among 7 die. By stars and bars, we have <math>\binom{3+7-1}{3}=\binom{9}{3}=\boxed{84}.</math> ~mathboy282 | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/HVn1WV80ZIU | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == Video Solution 2== | ||
+ | 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 12:12, 19 February 2021
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
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 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.
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.
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
Solution 5 (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 6 (Solution 5 but more clearer and compact)
Assume each die has value 1. Then we have left. This is to be split among 7 die. By stars and bars, we have ~mathboy282
Video Solution 1
~savannahsolver
Video Solution 2
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.