1988 AMC 12 第 18 题

先试着解答 1988 AMC 12 第 18 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 1988 AMC 12 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。

18.

一项职业保龄球锦标赛结束时,排名前 55 的选手进行附加赛。首先,编号 55 与编号 44 比赛。负者获得第 55 名奖金,胜者再与编号 33 比赛。这场比赛的负者获得第 44 名奖金,胜者再与编号 22 比赛。这场比赛的负者获得第 33 名奖金,胜者再与编号 11 比赛。最后一场的胜者获得第 11 名奖金,负者获得第 22 名奖金。编号 11 到编号 55 的选手共有多少种获奖名次顺序?
At the end of a professional bowling tournament, the top 55 bowlers have a play-off. First #55 bowls #4.4. The loser receives 55th prize and the winner bowls #33 in another game. The loser of this game receives 44th prize and the winner bowls #2.2. The loser of this game receives 33rd prize and the winner bowls #1.1. The winner of this game gets 11st prize and the loser gets 22nd prize. In how many orders can bowlers #11 through #55 receive the prizes?
1010
1616
2424
120120
以上都不是
none of these
答案:B
知识点:countingbinary choicestournament outcomes
小提示:
一共进行四场比赛
Exactly four games are played
大提示:
每一种胜者序列都确定一种不同的获奖名次顺序
Each sequence of winners determines a different prize order
解答:
四场比赛各有两种可能的胜者。四场比赛的结果序列唯一确定五个获奖名次,所以共有 24=162^4=16 种顺序。
所以正确答案是 B。
Each of the four games has two possible winners. A sequence of four outcomes uniquely determines the five prize positions, so there are 24=162^4=16 orders.
Thus the correct answer is B.
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