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2000 AMC 10 Problem 13

Problem 13 of 25IntermediateCounting & Probability

There are 55 yellow pegs, 44 red pegs, 33 green pegs, 22 blue pegs, and 11 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

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Solution

The board has five rows and five columns. To avoid two yellow pegs in a row or column, there must be exactly one yellow peg in each row, forcing the yellow pegs onto the long diagonal. The four red pegs must then each go in rows 22 through 5,5, and the only positions left force them into a single diagonal as well. Continuing with green, blue, and orange, every color is forced into a unique position. Hence there is exactly one valid arrangement. Thus, the correct answer is B.

More practice

Concepts: arrangements with restrictions · logical deduction

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