2024 AMC 12A Problems/Problem 16

A set of $12$ tokens ---- $3$ red, $2$ white, $1$ blue, and $6$ black ---- is to be distributed at random to $3$ game players, $4$ 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 $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? $\textbf{(A) }387 \qquad \textbf{(B) }388 \qquad \textbf{(C) }389 \qquad \textbf{(D) }390 \qquad \textbf{(E) }391 \qquad$

Solution 1 (12fact bash)

We have $12!$ total possible arrangements of $12$ distinct tokens. If we imagine the first $4$ tokens of our arrangement go to the first player, the next $4$ go to the second, and the final $4$ 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 $3!$ , $2!$ , $1!$ , and $6!$ to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.

Letting $R$ denote red, $W$ denote white, $B$ denote blue, and $L$ denote black, then our arrangement must be something like $RRRLWWLLWLLL$ . The three players are arbitrary, so we multiply by $3!$ ; then, the player who gets the reds has $\dbinom41=4$ possible arrangements, the player who gets the whites has $\dbinom42=6$ possibilities, and the player who gets the blacks has $\dbinom43=4$ possibilities. Our total on top is thus $3!\cdot2!\cdot1!\cdot6!\cdot3!\cdot4\cdot6\cdot4$ , and the denominator is $12!$ . Firstly, we have the $6!$ in the numerator cancel out part of the denominator; we thus have the following: