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2007 AMC 10B Problem 20

Problem 20 of 25HarderCounting & Probability

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?

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Solution

Choose 33 of the 55 rows in (53)=10\binom{5}{3}=10 ways and 33 of the 55 columns in (53)=10\binom{5}{3}=10 ways. The three chosen blocks must occupy distinct rows and columns, so they form a matching between the three rows and three columns, which can be done in 3!=63!=6 ways. The total is 10106=600.10\cdot 10\cdot 6=600. Thus, the correct answer is C.

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Concepts: combinations · permutations · arrangements with restrictions

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.