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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+ | 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. | ||
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+ | In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2017|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:55, 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?
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
Case 3. <RA>=1, and
In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.
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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