Question

There are $5$ yellow pegs, $4$ red pegs, $3$ green pegs, $2$ blue pegs, and $1$ 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?

$0$

$1$

$5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!$

$15!/(5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!)$

$15!$

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 $2$ through $5,$ 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.

2000 AMC 10 Problem 13

Solution:

Problem 13 in Other Years