A penny is placed in the coordinate plane $\left(0,0\right)$. The penny can be moved $1$ unit to the right, $1$ unit up, or diagonally $1$ unit to the right and $1$ unit up. How many different ways are there for the penny to get to the point $\left(5,5\right)$?

$\text{(A) }8\qquad\text{(B) }25\qquad\text{(C) }99\qquad\text{(D) }260\qquad\text{(E) }351$
1. **Initialization:**
   - \(dp[0][0] = 1\) because there is exactly one way to be at the starting point \((0,0)\).

2. **Recurrence Relation:**
   - For each point \((i,j)\), the number of ways to get there is the sum of the number of ways to get to the points \((i-1,j)\), \((i,j-1)\), and \((i-1,j-1)\) (if these points are within bounds).
   - Mathematically, this can be written as:
     \[
     dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1]
     \]
   - If \(i = 0\) or \(j = 0\), we need to handle the boundary conditions:
     \[
     dp[i][j] = dp[i-1][j] + dp[i][j-1] \quad \text{if } i = 0 \text{ or } j = 0
     \]

3. **Filling the DP Table:**
   - We will fill the table from \((0,0)\) to \((5,5)\) using the recurrence relation.

### Calculation:

Let's fill the DP table step-by-step:

\[
\begin{array}{c|cccccc}
i \backslash j & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
0 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 3 & 5 & 7 & 9 & 11 \\
2 & 1 & 5 & 13 & 25 & 41 & 61 \\
3 & 1 & 7 & 25 & 63 & 129 & 231 \\
4 & 1 & 9 & 41 & 129 & 321 & 681 \\
5 & 1 & 11 & 61 & 231 & 681 & 1573 \\
\end{array}
\]

- The value at \(dp[5][5]\) is 1573.

The final answer is \(\boxed{1573}\)