2004 AMC 8 第 4 题

先试着解答 2004 AMC 8 第 4 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2004 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

4.

汉密尔顿老师的八年级班想参加一年一度的三人篮球队锦标赛。兰斯、萨莉、乔伊和弗雷德被选为队员。三名首发队员有多少种选择方式?
Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
22
44
66
88
1010
答案:B
知识点:组合补集计数
小提示:
选择三名首发等同于选择谁不首发
Choosing the three starters is the same as choosing who sits out.
大提示:
有 44 名可能不首发的队员
There are 44 possible players who could be the alternate.
解答:
若有 33 名首发,则有一人不首发。选择不首发的人就确定了首发。
不首发的人有 44 种选择。
所以正确答案是 B。
If there are 33 starters, then one person must not be starting. Choosing the person who doesn’t start determines the starters.
There are 44 choices for the person who doesn’t start.
Thus, B is the correct answer.
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