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2007 AMC 10B Problem 20

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2007 AMC 10B Problem 20AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 3 of 5 · about 3 min · position P16-20

A set of 2525 square blocks is arranged into a 5×55 \times 5 square. How many different combinations of 33 blocks can be selected from that set so that no two are in the same row or column?

  1. A)

    100100

  2. B)

    125125

  3. C)

    600600

  4. D)

    23002300

  5. E)

    36003600

Answer

C

Key insight

Pick the 3 rows and 3 columns (10 ways each), then match rows to columns in 3! ways: 10 x 10 x 6 = 600.

Solution

A valid set of three blocks uses three distinct rows and three distinct columns, and the set is determined by which rows, which columns, and how they are paired up.

  • Choose the 33 rows: (53)=10\binom{5}{3} = 10 ways.
  • Choose the 33 columns: (53)=10\binom{5}{3} = 10 ways.
  • Assign each chosen row to a different chosen column: 3!=63! = 6 ways.

Every combination of blocks arises from exactly one such choice, so the count is
10⋅10⋅6=600.10 \cdot 10 \cdot 6 = 600.

The answer is (C) 600\boxed{\textbf{(C)}\ 600}.

Why this works

"No two share a row or column" is a partial permutation: each block occupies its own row and column, like non-attacking rooks. Separating the which rows/columns decision from the how they pair decision keeps the count unordered from the start, avoiding a division by 3!3! at the end.

Alternative approach

Pick blocks one at a time: 2525 choices for the first, then 1616 (avoiding its row and column), then 99. That counts ordered triples, 25⋅16⋅9=360025 \cdot 16 \cdot 9 = 3600, so divide by 3!=63! = 6 to get 600600.

The trap

Counting ordered selections (25 x 16 x 9 = 3600) and forgetting to divide by 3! for the unordered combination.

Common mistakes

  • Counting ordered selections (25 x 16 x 9 = 3600) and forgetting to divide by 3! for the unordered combination.
  • Choosing rows and columns but omitting the 3!3! pairings, which gives 100100.

Techniques

Map the objects to something easier to count · Set up the equation/formula and compute; no special trick needed