2000 AMC 10 Question 13 - thichhoctoan.com

Question 13

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?

Question illustration

$0$
$1$
$5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!$
$\frac{15!}{5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!}$
$15!$