2024 AMC 10B Problem 17

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

Answer: D

Concepts:permutations combinations casework

Difficulty rating: 1730

Small hint:

Count results with no tie separately from results with exactly one tied group

Big hint:

For a tied group of

k

snails, choose them in

\binom{5}{k}

ways and order the

6 - k

resulting blocks in

(6-k)!

ways

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.