How many finite sequences $x_1,x_2,\ldots,x_m$ are there such that $x_i=1$ or $2$ and $\sum_{i=1}^mx_i=10$?
$\textbf{(A)}~89$
$\textbf{(B)}~73$
$\textbf{(C)}~107$
$\textbf{(D)}~119$
1. **Define the problem in terms of a recurrence relation:**
   We need to find the number of sequences \( x_1, x_2, \ldots, x_m \) such that \( x_i = 1 \) or \( 2 \) and \( \sum_{i=1}^m x_i = 10 \). This is equivalent to finding the number of ways to reach the 10th step if you can take either 1 step or 2 steps at a time.

2. **Establish the recurrence relation:**
   Let \( a_n \) be the number of ways to reach the \( n \)-th step. We can reach the \( n \)-th step either from the \( (n-1) \)-th step by taking 1 step or from the \( (n-2) \)-th step by taking 2 steps. Therefore, the recurrence relation is:
   \[
   a_n = a_{n-1} + a_{n-2}
   \]

3. **Determine the base cases:**
   - \( a_0 = 1 \): There is 1 way to stay at the ground (doing nothing).
   - \( a_1 = 1 \): There is 1 way to reach the first step (taking one 1-step).

4. **Compute the values using the recurrence relation:**
   \[
   \begin{aligned}
   a_2 &= a_1 + a_0 = 1 + 1 = 2 \\
   a_3 &= a_2 + a_1 = 2 + 1 = 3 \\
   a_4 &= a_3 + a_2 = 3 + 2 = 5 \\
   a_5 &= a_4 + a_3 = 5 + 3 = 8 \\
   a_6 &= a_5 + a_4 = 8 + 5 = 13 \\
   a_7 &= a_6 + a_5 = 13 + 8 = 21 \\
   a_8 &= a_7 + a_6 = 21 + 13 = 34 \\
   a_9 &= a_8 + a_7 = 34 + 21 = 55 \\
   a_{10} &= a_9 + a_8 = 55 + 34 = 89
   \end{aligned}
   \]

5. **Conclusion:**
   The number of finite sequences \( x_1, x_2, \ldots, x_m \) such that \( x_i = 1 \) or \( 2 \) and \( \sum_{i=1}^m x_i = 10 \) is \( \boxed{89} \).