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

(Solution)
m (solution page problem is different from problems page problem)
(46 intermediate revisions by 25 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>
 +
\textbf{(A) }42\qquad
 +
\textbf{(B) }49\qquad
 +
\textbf{(C) }56\qquad
 +
\textbf{(D) }63\qquad
 +
\textbf{(E) }84\qquad
 +
</math>
 +
 
 +
== 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.  
 
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, 1-3 dices' number on the top face must be increased by a total of 3.  
+
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:
 
There are 3 ways to do so:
 
3, 2+1, and 1+1+1
 
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.
+
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>
 +
 
 +
===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.
  
Therefore, the answer is <math>7+42+35 = \boxed {(E) 84}</math>
+
===Solution 3===
 +
Add possibilities. There are <math>3</math> ways to sum to <math>10</math>, listed below.
  
Solution by PancakeMonster2004
+
<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>.
 +
 
 +
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 ==
Line 21: Line 61:
 
{{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]]

Revision as of 12:12, 19 February 2021

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

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 $7$ for Case 1, $7\cdot 6 = 42$ for Case 2, and $\frac{7\cdot 6\cdot 5}{3!} = 35$ for Case 3.

Therefore, the answer is $7+42+35 = \boxed {\textbf{(E) } 84}$

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

Solution 3

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

Thus we have repeated Solution 1 exactly, but with less explanation.

Solution 4 (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 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 $10$ - $7$ = $3$ 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 $\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{84}$

Solution 6 (Solution 5 but more clearer and compact)

Assume each die has value 1. Then we have $10-(1 \cdot 7)=3$ left. This is to be split among 7 die. By stars and bars, we have $\binom{3+7-1}{3}=\binom{9}{3}=\boxed{84}.$ ~mathboy282

Video Solution 1

https://youtu.be/HVn1WV80ZIU

~savannahsolver

Video Solution 2

https://youtu.be/5UojVH4Cqqs?t=5381

~ pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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