2024 AMC 10B Problem 17

A result is an order of finish where one group of snails may share a place. Since the tie can have $2, 3, 4$ or $5$ snails, or there can be no tie at all, split into cases by the size of the tie.

1. No tie. The $5$ snails finish in $5$ different places: $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ results.
2. A tie of some size. First choose which snails tie. Then glue them together into one block, because tied snails have no order among themselves. Finally line up the block and the remaining snails.
3. Tie of $2$. Choose the $2$ tied snails: $\binom{5}{2} = 10$ ways. Now there are $4$ things to line up (the block and $3$ single snails): $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways. Total $10 \cdot 24 = 240$.
4. Tie of $3$. Choose them: $\binom{5}{3} = 10$ ways. Line up $3$ things: $6$ ways. Total $60$.
5. Tie of $4$. Choose them: $5$ ways. Line up $2$ things (the block first or the lone snail first): $2$ ways. Total $10$.
6. All $5$ tie. Just $1$ result.
7. Add the cases:
$$120 + 240 + 60 + 10 + 1 = 431.$$

The answer is $\boxed{\textbf{(D)}\ 431}$.

"At most one tie" means the result is a lineup of groups where only one group can hold more than one snail. Choosing the group, gluing it into a block and lining up the blocks is the standard way to count lineups with one glued group. The example in the problem (three snails tied for second) is a reminder that ties of any size count.

Use the answer choices as a check. The counts for no tie and for ties of $2$, $3$ and $4$ are all multiples of $10$, and the five-way tie adds $1$, so the total ends in $1$. That leaves $361$ and $431$. The first two cases alone give $120 + 240 = 360$, and the other cases add more than $1$, so the total is above $361$. That leaves $431$.