2013 AMC 10A 第 11 题

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：

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