2024 AMC 12A Problems/Problem 16

Problem

A set of tokens ---- red, white, blue, and black ---- is to be distributed at random to game players, 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 , where and are relatively prime positive integers. What is ?

Solution 1 (12fact bash)

We have total possible arrangements of distinct tokens. If we imagine the first tokens of our arrangement go to the first player, the next go to the second, and the final go to the third, then we can view this problem as counting the number of valid arrangements.

Firstly, the tokens are not all distinct, so we multiply by , , , and to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.

Letting denote red, denote white, denote blue, and denote black, then our arrangement must be something like . The three players are arbitrary, so we multiply by ; then, the player who gets the reds has possible arrangements, the player who gets the whites has possibilities, and the player who gets the blacks has possibilities. Our total on top is thus , and the denominator is . Firstly, we have the in the numerator cancel out part of the denominator; we thus have the following:

Our answer is

~Technodoggo

Solution 2

Assume all of them are distinct even when some have the same color, Total possibility Next, assume that all the token are already in 3 different groups (3! Ways to do so) We then split the 6 black token into these 3 different groups There are a total of 3!(6c3)(3c2)(1c1) ways in doing so This the answer 3!(6!/1!2!3!)/(12!/4!4!4!)=4/385 So the answer is

See also

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