2021 Fall AMC 12A Problems/Problem 18

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 are indistinguishable and the bins are distinguishable.

Let $d$ be the number of ways to distribute $20$ balls into $5$ bins. We have $p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.$ Therefore, the answer is $\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.$

Remark

By the stars and bars argument, we get $d=\binom{20+5-1}{5-1}=\binom{24}{4}.$

~MRENTHUSIASM

Solution 2 (Binomial Coefficients)

For simplicity purposes, the balls are indistinguishable and the bins are distinguishable.

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

Notice that 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. We have $x \cdot \frac{\binom{5}{1} \cdot \binom{4}{1} \cdot \binom{4}{1}}{5} = 16x.$ Therefore, we get $p = \frac{16x}{a},$ 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.

From any element in $A$ , we may take 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, then $5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}$ .

Solution 5

Denote by $\left( x_1, \cdots, x_{20} \right)$ as an outcome, where $x_i \in \left\{ 1 , \cdots , 5 \right\}$ denotes the index of a bin that ball $i$ is dropped into.

Denote by $\Omega$ the sample space.

Denote by $E_p$ the event that defines $p$ .

We compute $| E_p |$ in the following steps.

Step 1: We determine which bin has 3 balls.

The number of ways is 5.

Step 2: We determine which bin has 5 balls.

The number of ways is 4.

Step 3: We allocate 20 balls into 5 bins with their designated number of balls

The number of ways is $\binom{20}{3, 5, 4, 4, 4}$ .

Putting all steps together, following from the rule of product, we have $| E_p | = 5 \cdot 4 \cdot \binom{20}{3, 5, 4, 4, 4}$ .

Denote by $E_q$ the event that defines $q$ .

Therefore, $| E_q | = \binom{20}{4, 4, 4, 4, 4}$ .

Therefore, $\begin{align*} \frac{p}{q} & = \frac{\frac{| E_p |}{| \Omega|}}{\frac{| E_q |}{| \Omega |}} \\ & = \frac{| E_p |}{| E_q |} \\ & = \frac{5 \cdot 4 \cdot \binom{20}{3, 5, 4, 4, 4}}{\binom{20}{4, 4, 4, 4, 4}} \\ & = \frac{5 \cdot 4 \cdot \frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}} \\ & = \frac{5 \cdot 4 \cdot 4! \cdot 4!}{3! \cdot 5!} \\ & = 16. \end{align*}$

Therefore, the answer is $\boxed{\textbf{(E) }16}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution by Mathematical Dexterity

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

Video Solution by Punxsutawney Phil

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

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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