Question

学生会要从成员中选出一个两人迎宾委员会和一个三人策划委员会。选出两人迎宾委员会恰好有 $10$ 种方法。学生可以同时在两个委员会中任职。三人策划委员会有多少种不同选法？

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?

$10$

$12$

$15$

$18$

$25$

答案：A

知识点：组合二次方程

难度评级：1020

小提示：

设成员数为 $n$ ，并使用 $\binom n2=10$ 。

Let $n$ be the number of council members and use $\binom n2=10$

大提示：

求出 $n$ 后，计算 $\binom n3$ 。

After finding $n$ , count $\binom n3$ planning committees

解答：

设学生会有 $x$ 名成员。选二人委员会的方法数为 $\binom{x}{2} = \dfrac{x(x - 1)}{2}\text{。}$ 这个值等于 $10$ ，所以 $x^2 - x = 20$ $x^2 - x - 20 = 0\text{。}$ 因式分解得 $(x - 5)(x + 4) = 0$ $x = 5\text{，}$ 因为学生人数不能为负数。

于是选 $3$ 人策划委员会的方法数为 $\binom{5}{3} = \binom{5}{2} = 10\text{。}$

所以正确答案是 A。

Let $x$ be the number of students. Then the number of ways to pick a two-person committee is $\binom{x}{2} = \dfrac{x(x - 1)}{2}.$ We know that this equals $10,$ so $x^2 - x = 20$ $x^2 - x - 20 = 0.$ Factoring yields $(x - 5)(x + 4) = 0$ $x = 5,$ since there cannot be a negative number of students.

Then, the number of ways to pick a $3$ -person committee is $\binom{5}{3} = \binom{5}{2} = 10.$

Thus, A is the correct answer.

2013 AMC 10A 第 11 题

答案：A

难度评级：1020

小提示：

大提示：

解答：

其他年份的第 11 题