AMC12 2024 A

AMC12 2024 A · Q16

AMC12 2024 A · Q16. It mainly tests Basic counting (rules of product/sum), Combinations.

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 $m+n$?

有12个令牌——3个红色的，2个白色的，1个蓝色的，6个黑色的——将随机分给3名游戏玩家，每人4个令牌。某玩家得到所有红色令牌，另一玩家得到所有白色令牌，剩下玩家得到蓝色令牌的概率可以写成 $\frac{m}{n}$，其中 $m$ 和 $n$ 是互质的正整数。求 $m+n$？

(A) 387 387

(B) 388 388

(D) 390 390

(E) 391 \qquad 391 \qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

We have $\binom{12}{4,4,4}$ ways to handle the red/white/blue tokens distribution on the denominator. Now we simply $\binom{6}{1}$ $\binom{5}{2}$ $3!$ for the numerator in order to handle the black tokens and distinguishable persons. The solution is therefore $\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}$ or $4+385=\boxed{\textbf{(C) }389}.$

分母上有 $\binom{12}{4,4,4}$ 种处理红/白/蓝令牌分配的方式。现在分子上简单地用 $\binom{6}{1}$ $\binom{5}{2}$ $3!$ 来处理黑色令牌和可区分的人。因此解为 $\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}$，或 $4+385=\boxed{\textbf{(C) }389}$。

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