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Difference between revisions of "2017 AMC 12B Problems/Problem 22"

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<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math>
 
<math>\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}</math>
  
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==Solution==
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==Solution 1==
 
It amounts to filling in a <math>4 \times 4</math> matrix. Columns <math>C_1 - C_4</math> are the random draws each round; rowof each player. Also, let <math>\%R_A</math> be the number of nonzero elements in <math>R_A</math>.
 
It amounts to filling in a <math>4 \times 4</math> matrix. Columns <math>C_1 - C_4</math> are the random draws each round; rowof each player. Also, let <math>\%R_A</math> be the number of nonzero elements in <math>R_A</math>.
 
Sidenote: (Not the author)(What does -1, 1, and 0, and R notation represent)?
 
Sidenote: (Not the author)(What does -1, 1, and 0, and R notation represent)?
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There's a grand total of <math>45</math> ways for this to happen, along with <math>12^3</math> total cases. The probability we're asking for is thus <math>\frac{45}{(12^3)}= \boxed{\textbf{(B)}\frac{5}{192}}</math>.
 
There's a grand total of <math>45</math> ways for this to happen, along with <math>12^3</math> total cases. The probability we're asking for is thus <math>\frac{45}{(12^3)}= \boxed{\textbf{(B)}\frac{5}{192}}</math>.
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LaTex polished by [[Grace.yongqing.yu]].
 
  
 
==Solution 2 (Less Casework)==
 
==Solution 2 (Less Casework)==

Revision as of 19:30, 15 November 2022

Problem

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

$\textbf{(A)}\quad \dfrac{7}{576} \qquad \qquad \textbf{(B)}\quad \dfrac{5}{192} \qquad\qquad \textbf{(C)}\quad \dfrac{1}{36} \qquad\qquad \textbf{(D)}\quad \dfrac{5}{144} \qquad\qquad\textbf{(E)}\quad \dfrac{7}{48}$

Solution 1

It amounts to filling in a $4 \times 4$ matrix. Columns $C_1 - C_4$ are the random draws each round; rowof each player. Also, let $\%R_A$ be the number of nonzero elements in $R_A$. Sidenote: (Not the author)(What does -1, 1, and 0, and R notation represent)?

WLOG, let $C_1 = \begin{pmatrix} 1\\-1\\0\\0\end{pmatrix}$. Parity demands that $\%R_A$ and $\%R_B$ must equal $2$ or $4$.

Case 1: $\%R_A = 4$ and $\%R_B = 4$. There are $\binom{3}{2}=3$ ways to place $2$ $-1$'s in $R_A$, so there are $3$ ways.

Case 2: $\%R_A = 2$ and $\%R_B=4$. There are $3$ ways to place the $-1$ in $R_A$, $2$ ways to place the remaining $-1$ in $R_B$ (just don't put it under the $-1$ on top of it!), and $2$ ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of $\%R_A = 4$, $\%R_B = 2$ for a total of $24$ ways.

Case 3: $\%R_A=\%R_B=2$. There are three ways to place the $-1$ in $R_A$. Now, there are two cases as to what happens next.

Sub-case 3.1: The $1$ in $R_B$ goes directly under the $-1$ in $R_A$. There's obviously $1$ way for that to happen. Then, there are $2$ ways to permute the two pairs of $1, -1$ in $R_C$ and $R_D$. (Either the $1$ comes first in $R_C$ or the $1$ comes first in $R_D$.)

Sub-case 3.2: The $1$ in $R_B$ doesn't go directly under the $-1$ in $R_A$. There are $2$ ways to place the $1$, and $2$ ways to do the same permutation as in Sub-case 3.1. Hence, there are $3(2+2 \cdot 2)=18$ ways for this case.

There's a grand total of $45$ ways for this to happen, along with $12^3$ total cases. The probability we're asking for is thus $\frac{45}{(12^3)}= \boxed{\textbf{(B)}\frac{5}{192}}$.

Solution 2 (Less Casework)

We will proceed by taking cases based on how many people are taking part in this "transaction." We can have $2$, $3$, or $4$ people all giving/receiving coins during the $4$ turns. Basically, (like the previous solution), we think of this as filling out a $4\text{x}2$ matrix of letters, where a letter on the left column represents this person gave, and a letter on the right column means this person received. We need to make sure that for each person that gave in total certain amount, they received in total from other people that same amount, or in other words, we want it such that there are an equal number of A's, B's, C's, and D's in both columns of the matrix.

For example, the matrix below represents: A gives B a coin, then B gives C a coin, then C gives D a coin, and finally A gives D a coin, in this order. \[ \begin{bmatrix} & A & B & \\ & B & C & \\ & C & D & \\ & A & D & \\ \end{bmatrix} \] Case 1: $2$ people. In this case, we have $\binom{4}{2}$ ways to choose the two people, and $\binom{4}{2}$ ways to order them to get a count of ${\binom{4}{2}}^2 = 36$ ways.

Case 2: $3$ people. In this case, one special person is giving/receiving twice. There are four ways to choose this person, then of the remaining three people we choose two, to be the people interacting with the special person. Thus, we have $4\cdot\binom{3}{2}\cdot4! = 288$ ways here.

Case 3: $4$ people. In this case. First note that no person can give/receive twice. If we keep the order of A, B, C, D giving in that order (and permute afterward), then there are three options to choose A's receiver, and three options for B's receiver afterward. Then it is uniquely determined who C and D give to. This gives a total of $3\cdot3\cdot4! = 216$ ways, after permuting.


So we have a total of $36+288+216=540$ ways to order the four pairs of people. Now we divide this by the total number of ways: $(4\cdot3)^4$ (four rounds, four ways to choose giver, three to choose receiver each round). So the answer is $\frac{540}{12^4} = \boxed{\textbf{(B)}\frac{5}{192}}$.

~ccx09

Latex polished by Argonauts16 and by Grace.yongqing.yu.

Solution 3

Similar to solution 2, we think in terms of transactions. WLOG, for the 1st transaction, we assume that A gives to B, which we denote AB. For the 2nd transaction, there are 12 options to choose from, so there are $12^3$ possible options.

Case 1 (Giving back immediately): The 2nd transaction is BA (1 option). Then, the 3rd transaction can be whatever you like (12 options), but the 4th transaction is now fixed to be the opposite of the 3rd transaction (1 option). So here we have $1\cdot 12\cdot 1 = 12$ good options.

Case 2 (Allows a cycle or two back-and-forths): The 2nd transaction is one of BC, BD, CA, DA, CD, DC (6 options). Then, for the 3rd transaction, 2 options force a "cycle" on the 4th transaction (example 2.1), and 2 options force two "back-and-forths" on the 4th transaction (example 2.2). In total, there are $2+2=4$ options for the 3rd transaction. So here we have $6\cdot 4\cdot 1 = 24$ good options.

Example 2.1: suppose 2nd = BC. Then if 3rd = DA or CD, the 4th is forced to be CD or DA respectively, completing a transaction "cycle"

Example 2.2: suppose 2nd = BC. Then if 3rd = AB or BC, the 4th is forced to be BC or AB respectively. In the end, both A-B and B-C had back-and-forth transactions

Case 3 (Allows two back-and-forths only): The 2nd transaction is one of AC, AD, DB, CB (4 options). Then, for the 3th transaction, there are only 2 possible options (namely, to reverse one of the previous transactions). Of course, the final transaction is forced. Here, we have $4\cdot 2\cdot 1 = 8$ good options.

Case 4 (AB again): The 2nd transaction is AB (same as 1st). This forces the later transactions to both be BA. Here we have 1 good option.

Summing, we have $12+24+8+1=45$ good options among $12^3$ total options, so the solution is $\frac{45}{(12^3)}= \boxed{\textbf{(B)}\frac{5}{192}}$.

SilverLion

Solution 4 Visualization of three cases

https://www.happymatheducation.com/uploads/1/3/6/4/136474762/2017amc12b22.jpg

~Di

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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