There are $5$ yellow pegs, $4$ red pegs, $3$ green pegs, $2$ blue pegs, and $1$ orange peg 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]

$\text{(A)}\ 0\qquad\text{(B)}\ 1\qquad\text{(C)}\ 5!\cdot4!\cdot3!\cdot2!\cdot1!\qquad\text{(D)}\ \frac{15!}{5!\cdot4!\cdot3!\cdot2!\cdot1!}\qquad\text{(E)}\ 15!$
1. **Understanding the Problem:**
   We need to place pegs of different colors on a triangular peg board such that no horizontal row or vertical column contains two pegs of the same color. The peg board has 15 positions arranged in a triangular shape.

2. **Analyzing the Peg Board:**
   The peg board can be visualized as follows:
   ```
   (0,4)
   (0,3) (1,3)
   (0,2) (1,2) (2,2)
   (0,1) (1,1) (2,1) (3,1)
   (0,0) (1,0) (2,0) (3,0) (4,0)
   ```
   Each row and column must contain pegs of different colors.

3. **Color Distribution:**
   - 5 yellow pegs
   - 4 red pegs
   - 3 green pegs
   - 2 blue pegs
   - 1 orange peg

4. **Placing the Pegs:**
   We need to place the pegs such that no two pegs of the same color are in the same row or column. The only possible arrangement that satisfies this condition is:
   ```
   Y
   RY
   GRY
   BGRY
   OBGRY
   ```
   This arrangement ensures that each row and each column contains pegs of different colors.

5. **Counting the Arrangements:**
   - There are \(5!\) ways to arrange the yellow pegs.
   - There are \(4!\) ways to arrange the red pegs.
   - There are \(3!\) ways to arrange the green pegs.
   - There are \(2!\) ways to arrange the blue pegs.
   - There is \(1!\) way to arrange the orange peg.

6. **Calculating the Total Number of Arrangements:**
   \[
   5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!
   \]
   \[
   = 120 \cdot 24 \cdot 6 \cdot 2 \cdot 1
   \]
   \[
   = 120 \cdot 24 \cdot 12
   \]
   \[
   = 120 \cdot 288
   \]
   \[
   = 34560
   \]

The final answer is \(\boxed{34560}\).