AMC12 2024 B

AMC12 2024 B · Q16

AMC12 2024 B · Q16. It mainly tests Permutations, Combinations.

A group of $16$ people will be partitioned into $4$ indistinguishable $4$-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^{r}M$, where $r$ and $M$ are positive integers and $M$ is not divisible by $3$. What is $r$?

有 $16$ 个人将被分成 $4$ 个不可区分的 $4$ 人委员会。每个委员会将有一位主席和一位秘书。进行这些分配的不同方式的数量可以写成 $3^{r}M$，其中 $r$ 和 $M$ 是正整数，且 $M$ 不可被 $3$ 整除。$r$ 是多少？

(A) 5 5

(B) 6 6

(D) 8 8

(E) 9 \qquad 9 \qquad

Answer

Correct choice: (A)

正确答案：（A）

Solution

There are ${16 \choose 4}$ ways to choose the first committee, ${12 \choose 4}$ ways to choose the second, ${8 \choose 4}$ for the third, and ${4 \choose 4}=1$ for the fourth. Since the committees are indistinguishable, we need to divide the product by $4!$. Thus the $16$ people can be grouped in \[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\] ways. In each committee, there are $4 \cdot 3=12$ ways to choose the chairperson and secretary, so $12^4$ ways for all $4$ committees. Therefore, there are \[\frac{16!}{(4!)^5}12^4\] total possibilities. Since $16!$ contains $6$ factors of $3$, $(4!)^5$ contains $5$, and $12^4$ contains $4$, $r=6-5+4=\boxed{\textbf{(A) }5}$.

选择第一个委员会有 ${16 \choose 4}$ 种方式，选择第二个有 ${12 \choose 4}$ 种，选择第三个有 ${8 \choose 4}$ 种，选择第四个有 ${4 \choose 4}=1$ 种。由于委员会不可区分，需要除以 $4!$。因此，将 $16$ 个人分组的方式数为 \[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\] 在每个委员会中，选择主席和秘书有 $4 \cdot 3=12$ 种方式，因此 $4$ 个委员会共有 $12^4$ 种方式。因此，总的可能性为 \[\frac{16!}{(4!)^5}12^4\] 由于 $16!$ 包含 $6$ 个 $3$ 的因子，$(4!)^5$ 包含 $5$ 个，$12^4$ 包含 $4$ 个，因此 $r=6-5+4=\boxed{\textbf{(A) }5}$。

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