Question

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 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?

$180$

$361$

$420$

$431$

$720$

Solution:

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

2024 AMC 10B Problem 17

Solution:

Problem 17 in Other Years