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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 18"

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~MRENTHUSIASM ~Jesshuang
 
~MRENTHUSIASM ~Jesshuang
  
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==Solution 2 (Binomial Coefficients)==
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==Solution 2 ==
 
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
 
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
  
 
Let <math>q=\frac{x}{a},</math> where <math>a</math> is the total number of combinations and <math>x</math> is the number of cases where every bin ends up with <math>4</math> balls.
 
Let <math>q=\frac{x}{a},</math> where <math>a</math> is the total number of combinations and <math>x</math> is the number of cases where every bin ends up with <math>4</math> balls.
  
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Notice that we can take <math>1</math> ball from one bin and place it in another bin so that some bin ends up with <math>3</math> balls, another with <math>5</math> balls, and the other three with <math>4</math> balls each. We have <cmath>p = \frac{x \cdot \binom41 \cdot \binom41}{a} = \frac{16x}{a},</cmath> from which <math>\frac{p}{q} = \boxed{\textbf{(E)}\ 16}.</math>
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We can take <math>1</math> ball from one bin and place it in another bin so that some bin ends up with <math>3</math> balls, another with <math>5</math> balls, and the other three with <math>4</math> balls each. Note that one configuration of <math>4{-}4{-}4{-}4{-}4</math> corresponds to <math>5\cdot4\cdot4=80</math> configurations of <math>3{-}5{-}4{-}4{-}4.</math> On the other hand, one configuration of <math>3{-}5{-}4{-}4{-}4</math> corresponds to <math>5</math> configurations of <math>4{-}4{-}4{-}4{-}4.</math>
 +
 
 +
Therefore, we have <cmath>p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},</cmath> from which <math>\frac{p}{q} = \boxed{\textbf{(E)}\ 16}.</math>
  
 
~Hoju
 
~Hoju
  
 
==Solution 3 (Binomial Coefficients) ==
 
==Solution 3 (Binomial Coefficients) ==
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Since both of the boxes will have <math>3</math> boxes with <math>4</math> balls in them, we can leave those out. There are <math>\binom {6}{3} = 20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the boxes. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4} = 70</math> ways to put the <math>4</math> balls inside the boxes. Therefore, we have <math>\frac {56\cdot 20}{70}</math> which is equal to <math>8 \cdot 2 = \boxed{\textbf{(E)}\ 16}</math>.
+
Since both of the cases will have <math>3</math> bins with <math>4</math> balls in them, we can leave those out. There are <math>2 \cdot \binom {5}{2} = 20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the bins. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4} = 70</math> ways to put the <math>4</math> balls inside the bins. Therefore, we have <math>\frac {56\cdot 20}{70}</math> which is equal to <math>8 \cdot 2 = \boxed{\textbf{(E)}\ 16}</math>.
  
 
~Arcticturn
 
~Arcticturn
  
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==Solution 4 (Set Theory) ==
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==Solution 4 (Set Theory / Graph Theory) ==
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Construct the set <math>A</math> consisting of all possible <math>3-5-4-4-4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4-4-4-4-4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>.  
+
Construct the set <math>A</math> consisting of all possible <math>3{-}5{-}4{-}4{-}4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4{-}4{-}4{-}4{-}4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>.
 +
 
 +
Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
 +
 
 +
For any element in <math>A</math>, we may choose one of the <math>5</math> balls in the <math>5</math>-bin and move it to the <math>3</math>-bin to get a valid element in <math>B</math>. This implies the number of edges is <math>5|A|</math>.  
  
−
Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> if it is possible to reach one configuration from the other by moving a single ball (note this process is reversible). Let us consider the total number of edges drawn.
+
On the other hand, for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math>-bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>.
  
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From any element in <math>A</math>, we may take one of the <math>5</math> balls in the <math>5</math>-bin and move it to the <math>3</math>-bin to get a valid element in <math>B</math>. This implies the number of edges is <math>5|A|</math>.  
+
We equate the expressions to get <math>5|A| = 80|B|</math>, from which <math>\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>.
  
−
On the other hand for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math> bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>.
+
== Video Solution by OmegaLearn ==
 +
https://youtu.be/mIJ8VMuuVvA?t=220
  
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Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>.
+
~ pi_is_3.14
  
 
==Video Solution by Mathematical Dexterity==
 
==Video Solution by Mathematical Dexterity==

Revision as of 11:12, 10 November 2023

The following problem is from both the 2021 Fall AMC 10A #21 and 2021 Fall AMC 12A #18, so both problems redirect to this page.

Problem

Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$

Solution 1 (Multinomial Coefficients)

For simplicity purposes, we assume that the balls and the bins are both distinguishable.

Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.\] ~MRENTHUSIASM ~Jesshuang

Solution 2

For simplicity purposes, we assume that the balls and the bins are both distinguishable.

Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.

We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$

Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{\textbf{(E)}\ 16}.$

~Hoju

Solution 3 (Binomial Coefficients)

Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$. After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{\textbf{(E)}\ 16}$.

~Arcticturn

Solution 4 (Set Theory / Graph Theory)

Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$.

Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.

For any element in $A$, we may choose one of the $5$ balls in the $5$-bin and move it to the $3$-bin to get a valid element in $B$. This implies the number of edges is $5|A|$.

On the other hand, for any element in $B$, we may choose one of the $20$ balls and move it to one of the other $4$-bins to get a valid element in $A$. This implies the number of edges is $80|B|$.

We equate the expressions to get $5|A| = 80|B|$, from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}$.

Video Solution by OmegaLearn

https://youtu.be/mIJ8VMuuVvA?t=220

~ pi_is_3.14

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=Lu6eSvY6RHE

Video Solution by Punxsutawney Phil

https://YouTube.com/watch?v=bvd2VjMxiZ4

Video Solution by TheBeautyofMath

https://youtu.be/TOSHQPb7vaM

~IceMatrix

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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