AMC10 2024 B

AMC10 2024 B · Q17

AMC10 2024 B · Q17. It mainly tests Basic counting (rules of product/sum), Casework.

In a race among $5$ snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?

在5只蜗牛的比赛中，最多有一个平局，但这个平局可以涉及任意数量的蜗牛。例如，结果可能是Dazzler第一；Abby、Cyrus和Elroy并列第二；Bruna第五。比赛可能有多少种不同的结果？

(A) 180 180

(B) 361 361

(D) 431 431

(E) 720 720

Answer

Correct choice: (D)

正确答案：（D）

Solution

Intuitive solution: We perform casework based on how many snails tie. Let's say we're dealing with the following snails: $A,B,C,D,E$. $5$ snails tied: All $5$ snails tied for $1$st place, so only $1$ way. $4$ snails tied: $A,B,C,D$ all tied, and $E$ either got $1$st or last. ${5}\choose{1}$ ways to choose who isn't involved in the tie and $2$ ways to choose if that snail gets first or last, so $10$ ways. $3$ snails tied: We have $ABC, D, E$. There are $3! = 6$ ways to determine the ranking of the $3$ groups. There are $5\choose2$ ways to determine the two snails not involved in the tie. So $6 \cdot 10 = 60$ ways. $2$ snails tied: We have $AB, C, D, E$. There are $4! = 24$ ways to determine the ranking of the $4$ groups. There are $5\choose{3}$ ways to determine the three snails not involved in the tie. So $24 \cdot 10 = 240$ ways. $1$ snail tied: This is basically just every snail for a place, so $5! = 120$ ways. The answer is $1+10+60+240+120 = \boxed{\text{(D) }431}$.

直观解法：根据平局蜗牛数量进行分类讨论。假设蜗牛为$A,B,C,D,E$。 $5$只蜗牛并列：全部5只蜗牛并列第一，只有$1$种方式。 $4$只蜗牛并列：$A,B,C,D$并列，$E$要么第一要么最后。选择谁不参与平局${5}\choose{1}$种方式，$E$第一或最后的$2$种方式，共$10$种。 $3$只蜗牛并列：有$ABC, D, E$。3个组的排名有$3! = 6$种方式。选择两个不参与平局的蜗牛${5}\choose{2}=10$种方式，共$6 \cdot 10 = 60$种。 $2$只蜗牛并列：有$AB, C, D, E$。4个组的排名有$4! = 24$种方式。选择三个不参与平局的蜗牛${5}\choose{3}=10$种方式，共$24 \cdot 10 = 240$种。 $1$只蜗牛（无平局）：就是每个蜗牛单独排名，$5! = 120$种方式。总计$1+10+60+240+120 = \boxed{\text{(D) }431}$。

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