2003 AMC 10A 第 21 题

先试着解答 2003 AMC 10A 第 21 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 2003 AMC 10A 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。

21.

帕特要从一个盘子中选择六块饼干,盘子里只有巧克力豆、燕麦和花生酱三种饼干,且每种至少有六块。可以选出多少种不同的六块饼干组合?
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?
2222
2525
2727
2828
729729
答案:D
知识点:隔板法
小提示:
数非负整数解 a+b+c=6a + b + c = 6 的个数。
Count the nonnegative integer solutions to a+b+c=6a + b + c = 6
大提示:
由插板法,个数是 (6+22)\dbinom{6 + 2}{2}。
By stars and bars this is (6+22)\dbinom{6 + 2}{2}
解答:
一种组合由三种饼干各取多少块决定,所以要数 a+b+c=6a + b + c = 6 的非负整数解。
由插板法,在 88 个位置中放 22 个隔板,个数为 (82)=28\dbinom{8}{2} = 28。
所以正确答案是 D。
An assortment is determined by how many of each type are chosen, so we count nonnegative integer solutions to a+b+c=6.a + b + c = 6.
By stars and bars, placing 22 dividers among 88 slots gives (82)=28\dbinom{8}{2} = 28 assortments.
Thus, the correct answer is D.
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