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

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==Solution==
 
==Solution==
It amountsI choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of <RA>=3 and <RB>=1. Total 24 subcases.
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It amounts to filling in a 4x4 matrux M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let <RA>=number of non-0 elements in RA.
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WOLG, let C1=[1,-1,0,0]. Parity demands that <RA> and <RB> = 1 or 3.
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CASE 1. <RA>=3, and <RB>=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here.
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Case 2. <RA>=1, and <RB>=3. 3 Choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of <RA>=3 and <RB>=1. Total 24 subcases.
  
 
Case 3. <RA>=1, and <RB>=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.
 
Case 3. <RA>=1, and <RB>=1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.

Revision as of 12:03, 17 February 2017

Problem 22

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

It amounts to filling in a 4x4 matrux M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let <RA>=number of non-0 elements in RA.

WOLG, let C1=[1,-1,0,0]. Parity demands that <RA> and = 1 or 3.

CASE 1. <RA>=3, and =3. 3 choices for -1in RA, then everything is determined. So 3 subcases here.

Case 2. <RA>=1, and =3. 3 Choices for the -1in RA, 2 choices for the rest of 1 -1 in RB, 2 choices between C and D to fill in the remaining pair of 1 and -1. Then double for symmetrical case of <RA>=3 and =1. Total 24 subcases.

Case 3. <RA>=1, and =1. 3 Choices for -1in RA. Then (1)1 in RB goes directly under 1 in RA. Then 2 choices to assign remaining two pairs of 1, -1in RC and RD. Or (2) 1 in RB goes not directly under 1 in RA: 2 choices here. Then 2 choices to finish. Hence 3(2+2×2)=18.

In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
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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