AMC10 2024 B

AMC10 2024 B · Q22

AMC10 2024 B · Q22. It mainly tests Basic counting (rules of product/sum), 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 $r$?

$16$人将被分成$4$个不可区分的$4$人委员会。每个委员会有一个主席和一个秘书。进行这些分配的不同方式数可以写成$3^{r}M$，其中$r$和$M$是正整数且$M$不被$3$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，第二个${12 \choose 4}$种，第三个${8 \choose 4}$种，第四个${4 \choose 4}=1$种。由于委员会不可区分，需要除以$4!$。因此，将$16$人分组的方式数为 \[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\] 每个委员会中，选择主席和秘书有$4 \cdot 3=12$种方式，因此$4$个委员会有$12^4$种方式。总方式数为 \[\frac{16!}{(4!)^5}12^4\] 由于$16!$含$3$的因数$6$个，$(4!)^5$含$5$个，$12^4$含$4$个，故$r=6-5+4=\boxed{\textbf{(A) }5}$。

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