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2024 AMC 10B Problem 17

Problem 17 of 25IntermediateCounting & Probability

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?

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Solution

If nobody ties, the 55 snails finish in 5!=1205! = 120 orders. Now allow exactly one tied group of size kk with 2≤k≤5.2 \le k \le 5. Choose the group in (5k)\binom{5}{k} ways, then treat it as one block, leaving 6−k6 - k blocks to arrange in (6−k)!(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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Concepts: permutations · combinations · casework

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.