AMC10 2024 A

AMC10 2024 A · Q9

AMC10 2024 A · Q9. It mainly tests Combinations, Inclusion–exclusion (basic).

In how many ways can $6$ juniors and $6$ seniors form $3$ disjoint teams of $4$ people so that each team has $2$ juniors and $2$ seniors?

$6$ 名低年级生和 $6$ 名高年级生可以组成多少种不同的 $3$ 个不相交的 $4$ 人团队，使得每个团队有 $2$ 名低年级生和 $2$ 名高年级生？

(A) 720 720

(B) 1350 1350

(D) 3280 3280

(E) 8100 8100

Answer

Correct choice: (B)

正确答案：（B）

Solution

The number of ways in which we can choose the juniors for the team are ${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$. Similarly, the number of ways to choose the seniors are the same, so the total is $90\cdot90=8100$. But we must divide the number of permutations of three teams, since the order in which the teams were chosen doesn't matter, which is $3!$. Thus, the answer is $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$

选择低年级生分到团队的方式有 ${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$ 种。同样，选择高年级生的方式相同，所以总共有 $90\cdot90=8100$ 种。但我们必须除以 $3!$，因为三个团队的顺序无关紧要。因此，答案是 $\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}$

Topics