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2024 AMC 10B Problem 17

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2024 AMC 10B Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Arrangements with Restrictions · difficulty 3 of 5 · about 4 min · position P16-20

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 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?

  1. A)

    180180

  2. B)

    361361

  3. C)

    420420

  4. D)

    431431

  5. E)

    720720

Answer

D

Key insight

Split by the size k of the single tie: choose the tied snails, then order the 6-k blocks: C(5,k)(6-k)!, plus 5! untied.

Solution

A result is an ordering of the five snails in which at most one group of snails shares a position. Classify by the size kk of the tied group.

  • No tie. All 55 finish in distinct places: 5!=1205! = 120.
  • Tie of kk snails (2≤k≤52 \le k \le 5). Choose the tied snails in (5k)\binom{5}{k} ways. Treat them as a single block; together with the other 5−k5 - k snails there are 6−k6 - k blocks to order, in (6−k)!(6-k)! ways.
  • k=2k = 2: (52)⋅4!=10⋅24=240\binom52 \cdot 4! = 10 \cdot 24 = 240
  • k=3k = 3: (53)⋅3!=10⋅6=60\binom53 \cdot 3! = 10 \cdot 6 = 60
  • k=4k = 4: (54)⋅2!=5⋅2=10\binom54 \cdot 2! = 5 \cdot 2 = 10
  • k=5k = 5: (55)⋅1!=1\binom55 \cdot 1! = 1

Total: 120+240+60+10+1=431120 + 240 + 60 + 10 + 1 = 431.

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

Why this works

"At most one tie" means the result is an ordering of blocks where exactly one block may have size greater than 11. Casework on the size of that block, followed by "choose the members, then permute the blocks," is the standard way to count arrangements with a single glued group. The example in the statement (a three-way tie for second) is a hint that ties of any size are allowed.

Alternative approach

Sanity check with the choices: the total must end in 11 because the no-tie, two-tie, three-tie and four-tie counts are all multiples of 1010 and the five-way tie adds 11. That leaves (B) 361361 and (D) 431431, and 120+240120 + 240 alone already exceeds 361361.

The trap

Forgetting the no-tie case (120) or the all-tied case (1), or ordering the tied snails among themselves as if a tie had an internal order.

Common mistakes

  • Forgetting the no-tie case (120) or the all-tied case (1), or ordering the tied snails among themselves as if a tie had an internal order.
  • Multiplying (5k)\binom{5}{k} by 5!5! or by (5−k)!(5-k)! instead of (6−k)!(6-k)!, which miscounts where the tied block can be placed.

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed