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2024 AMC 12A Problem 16

Problem 16 of 25IntermediateCombinatoricsProbability & Statistics

A set of 1212 tokens — 33 red, 22 white, 11 blue, and 66 black — is to be distributed at random to 33 game players, 44 tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m+n?

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Solution

Treat all tokens as distinct; the total number of ways to deal 44 to each player is (124,4,4)=34650.\binom{12}{4,4,4}=34650. For the favorable event, choose which player gets the reds, whites, and blue in 3!=63!=6 ways. The red player needs 11 more token, the white player 22 more, and the blue player 33 more, all black; the 66 black tokens split as 1,2,31,2,3 in 6!1! 2! 3!=60\tfrac{6!}{1!\,2!\,3!}=60 ways. So the probability is 6⋅6034650=36034650=4385.\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}. Then m+n=4+385=389.m+n=4+385=389. Thus, the correct answer is C.
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Tagged: basic probability · multiset permutations

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