2013 AMC 10A 第 11 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

11.

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 1010 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 1010 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?

1010

1212

1515

1818

2525

答案:A
知识点:组合二次方程
难度评级:1020
解答:

设学生会有 xx 名成员。选二人委员会的方法数为 这个值等于 1010,所以 因式分解得 因为学生人数不能为负数。 (x2)=x(x1)2. \binom{x}{2} = \dfrac{x(x - 1)}{2}. x2x=20 x^2 - x = 20 x2x20=0. x^2 - x - 20 = 0. (x5)(x+4)=0 (x - 5)(x + 4) = 0 x=5, x = 5,

所以成员数为五名。选 33 人策划的方式数为 (53)=(52)=10. \binom{5}{3} = \binom{5}{2} = 10.

所以正确答案是 A

Let xx be the number of students. Then the number of ways to pick a two-person committee is (x2)=x(x1)2. \binom{x}{2} = \dfrac{x(x - 1)}{2}. We know that this equals 10,10, so x2x=20 x^2 - x = 20 x2x20=0. x^2 - x - 20 = 0. Factoring yields (x5)(x+4)=0 (x - 5)(x + 4) = 0 x=5, x = 5, since there cannot be a negative number of students.

Then, the number of ways to pick a 33-person committee is (53)=(52)=10. \binom{5}{3} = \binom{5}{2} = 10.

Thus, A is the correct answer.

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