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

Revision as of 04:11, 31 July 2022

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 (Binomial Coefficients)

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 boxes will have $3$ boxes with $4$ balls in them, we can leave those out. There are $\binom {6}{3} = 20$ 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 boxes. 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 boxes. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{\textbf{(E)}\ 16}$ .

~Arcticturn

Solution 4 (Set 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 $80|B|$ .

Since they must be equal, we have $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}$ .

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

@@ Line 35: / Line 35: @@
 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.
-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>.
+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>.
-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>.
+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>.
-Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>.
+Since they must be equal, we have <math>5|A| = 80|B|</math>, from which <math>\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}</math>.
 ==Video Solution by Mathematical Dexterity==

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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