2024 AMC 10B Problem 17

Attempt Problem 17 of the 2024 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

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17.

In a race among 55 snails, there is at most one tie, but that tie can involve any number of snails. For example, the result of the race 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?

180180

361361

420420

431431

720720

Answer: D
Concepts:permutationscombinationscasework
Difficulty rating: 1730
Solution:

If nobody ties, the 55 snails finish in 5!=1205! = 120 orders. Now allow exactly one tied group of size kk with 2k5.2 \le k \le 5. Choose the group in (5k)\binom{5}{k} ways, then treat it as one block, leaving 6k6 - k blocks to arrange in (6k)!(6 - k)! ways. Summing over k:k: (52)4!\binom{5}{2}4! +(53)3!+ \binom{5}{3}3! +(54)2!+ \binom{5}{4}2! +(55)1!+ \binom{5}{5}1! =240+60+10+1= 240 + 60 + 10 + 1 =311.= 311. Add the no-tie count: 120+311=431.120 + 311 = 431. Thus, D is the correct answer.

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