Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 10A Problems/Problem 11"

(→‎Solution)
(→‎Solution 5 (Similar to above, using number separation))
 
(84 intermediate revisions by 36 users not shown)
Line 1: Line 1:
−
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 <cmath>\frac{n}{6^7},</cmath>where <math>n</math> is a positive integer. What is <math>n</math>?
+
==Problem==
  
−
<math>\textbf{(A) }   42  \qquad        \textbf{(B) }  49  \qquad    \textbf{(C) }   56  \qquad  \textbf{(D) } 63 \qquad  \textbf{(E) }  84 </math>
+
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>?
  
−
== Solution ==
+
<math>
−
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.
+
\textbf{(A) }42\qquad
 +
\textbf{(B) }49\qquad
 +
\textbf{(C) }56\qquad
 +
\textbf{(D) }63\qquad
 +
\textbf{(E) }84\qquad
 +
</math>
  
−
In order for the sum to be exactly 10, 1-3 dices' number on the top face must be increased by a total of 3.
+
== Solutions ==
 +
===Solution 1===
  
−
There are 3 ways to do so:
+
Add possibilities. There are <math>3</math> ways to sum to <math>10</math>, listed below.
−
3, 2+1, and 1+1+1
 
  
−
There are <math>\dbinom {7}{1}</math> for Case 1, <math>7*6 = 42</math> for Case 2, and <math>\dbinom {7}{3}</math> for Case 3.
+
<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>
  
−
Therefore, the answer is <math>7+42+35 = \boxed {(E) 84}</math>
+
Add up the possibilities: <math>35+42+7=\boxed{\textbf{(E) }  84}</math>.
  
−
Solution by PancakeMonster2004
+
===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 <math>\dbinom{9}{3}=\boxed{\textbf{(E) } 84}</math>
 +
 
 +
===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 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{\textbf{(E) }84}</math>
 +
 
 +
 
 +
 
 +
=== 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
 +
<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.
 +
<cmath>\begin{align*}
 +
D &= 10 \\
 +
a+b+c+d+e+f+g &= 10 \\
 +
\dbinom{10-1}{7-1} &= \\
 +
\dbinom{9}{6} &= \\
 +
\dbinom{9}{3} &= \\
 +
\dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} &= \\
 +
12 \cdot 7 &= \boxed{\textbf{(E) }84} \\
 +
\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.
 +
 
 +
~ Wiselion =)
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/gTpg8yInCCY
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==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 ==
Line 21: Line 71:
 
{{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 15:13, 13 October 2024

Problem

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 \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$?

$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$

Solutions

Solution 1

Add possibilities. There are $3$ ways to sum to $10$, listed below.

\[4,1,1,1,1,1,1: 7\] \[3,2,1,1,1,1,1: 42\] \[2,2,2,1,1,1,1: 35.\]

Add up the possibilities: $35+42+7=\boxed{\textbf{(E) } 84}$.

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 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 $\dbinom{9}{3}=\boxed{\textbf{(E) } 84}$

Solution 3 (overkill)

We can use generating functions, where $(x+x^2+...+x^6)$ is the function for each die. We want to find the coefficient of $x^{10}$ in $(x+x^2+...+x^6)^7$, which is the coefficient of $x^3$ in $\left(\frac{1-x^7}{1-x}\right)^7$. This evaluates to $\dbinom{-7}{3} \cdot (-1)^3=\boxed{\textbf{(E) }}84$

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 $10$ - $7$ = $3$ 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 $3$ balls into $7$ boxes. From there we get $\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{\textbf{(E) }84}$


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 $D\{a, b, c, d, e, f, g, h\}$, we can say $D = 10$ and each of $D$'s elements are larger than $0$. Using the positive number separation formula, which is $\dbinom{n-1}{r-1}$, we can make the following equations. \begin{align*} D &= 10 \\ a+b+c+d+e+f+g &= 10 \\ \dbinom{10-1}{7-1} &= \\ \dbinom{9}{6} &= \\ \dbinom{9}{3} &= \\ \dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} &= \\ 12 \cdot 7 &= \boxed{\textbf{(E) }84} \\ \end{align*}

Note: We are unable to use non-negative number separations due to the fact that the dice *must* be larger than $0$ or positive.

~ Wiselion =)

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/gTpg8yInCCY

~Education, the Study of Everything


Video Solution 1

https://youtu.be/HVn1WV80ZIU

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_11&oldid=229752"