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2010 AMC 10A Problem 18

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2010 AMC 10A Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 6 min · position P16-20

Bernardo randomly picks 3 distinct numbers from the set {1,2,3,4,5,6,7,8,9}\{1,2,3,4,5,6,7,8,9\} and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set {1,2,3,4,5,6,7,8}\{1,2,3,4,5,6,7,8\} and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

  1. A)

    4772\frac{47}{72}

  2. B)

    3756\frac{37}{56}

  3. C)

    23\frac{2}{3}

  4. D)

    4972\frac{49}{72}

  5. E)

    3956\frac{39}{56}

Answer

B

Key insight

If Bernardo draws a 9 (probability 1/3) he wins outright; otherwise both draw from the same set and he wins with probability (1 - 1/56)/2 by symmetry.

Solution

Each selection is an unordered set of three digits, so Bernardo has (93)=84\binom93 = 84 equally likely choices and Silvia has (83)=56\binom83 = 56.

Case 1: Bernardo picks a 9. There are (82)=28\binom82 = 28 such sets, probability 2884=13\dfrac{28}{84} = \dfrac13. His number then begins with 99, while Silvia's begins with at most 88, so Bernardo wins for sure.

Case 2: Bernardo does not pick a 9 (probability 23\dfrac23). Now both players choose uniformly from the same 5656 sets, so the situations are symmetric. They tie exactly when they pick the same set, probability 156\dfrac{1}{56}. The rest of the probability splits evenly, so Bernardo wins with probability
1−1562=55112.\frac{1 - \frac{1}{56}}{2} = \frac{55}{112}.

Combining:
13⋅1+23⋅55112=13+55168=56+55168=111168=3756.\frac13 \cdot 1 + \frac23 \cdot \frac{55}{112} = \frac13 + \frac{55}{168} = \frac{56 + 55}{168} = \frac{111}{168} = \frac{37}{56}.

The answer is (B) 3756\boxed{\textbf{(B)}\ \frac{37}{56}}.

Why this works

The descending arrangement means each number corresponds to one set of digits, so the sample spaces are sets, not sequences. The only asymmetry between the players is the digit 99; split on it, and the remaining situation is a symmetric contest where "win, lose, tie" must sum to 11 with win and lose equal. Symmetry plus the tie correction avoids any comparison of 84×5684 \times 56 pairs.

The trap

Forgetting the tie case in the symmetric situation and using 1/2 instead of 55/112, which gives 2/3 (choice C).

Common mistakes

  • Forgetting the tie case in the symmetric situation and using 1/2 instead of 55/112, which gives 2/3 (choice C).
  • Treating the digit choices as ordered (permutations) for one player but unordered for the other.

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects