# Difference between revisions of "2000 AMC 10 Problems/Problem 13"

## Problem

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?

$[asy] unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((0,3)); dot((1,3)); dot((0,4)); [/asy]$

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 5!\cdot 4!\cdot 3!\cdot 2!\cdot 1! \qquad\mathrm{(D)}\ \frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!} \qquad\mathrm{(E)}\ 15!$

## Solution

The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable.

Clearly, there is only 1 possible ordering if the colors are indistinguishable.

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

Or, C.