What is the coefficient of $x^5$ in the expansion of $(1 + x + x^2)^9$?

$ 
\textbf{a)}\ 1680
\qquad\textbf{b)}\ 882
\qquad\textbf{c)}\ 729
\qquad\textbf{d)}\ 450
\qquad\textbf{e)}\ 246
$
To find the coefficient of \(x^5\) in the expansion of \((1 + x + x^2)^9\), we can use the multinomial theorem. The multinomial theorem states that:

\[
(a_1 + a_2 + \cdots + a_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} \frac{n!}{k_1! k_2! \cdots k_m!} a_1^{k_1} a_2^{k_2} \cdots a_m^{k_m}
\]

In our case, \(a_1 = 1\), \(a_2 = x\), and \(a_3 = x^2\), and we need to find the coefficient of \(x^5\) in \((1 + x + x^2)^9\). We need to find all combinations of \(k_1\), \(k_2\), and \(k_3\) such that:

\[
k_1 + k_2 + k_3 = 9 \quad \text{and} \quad k_2 + 2k_3 = 5
\]

We will consider each possible combination of \(k_2\) and \(k_3\) that satisfies these equations.

1. **Case 1: \(k_2 = 5\) and \(k_3 = 0\)**
   \[
   k_1 + 5 + 0 = 9 \implies k_1 = 4
   \]
   The coefficient is:
   \[
   \frac{9!}{4!5!0!} = \frac{362880}{24 \cdot 120 \cdot 1} = 126
   \]

2. **Case 2: \(k_2 = 3\) and \(k_3 = 1\)**
   \[
   k_1 + 3 + 1 = 9 \implies k_1 = 5
   \]
   The coefficient is:
   \[
   \frac{9!}{5!3!1!} = \frac{362880}{120 \cdot 6 \cdot 1} = 504
   \]

3. **Case 3: \(k_2 = 1\) and \(k_3 = 2\)**
   \[
   k_1 + 1 + 2 = 9 \implies k_1 = 6
   \]
   The coefficient is:
   \[
   \frac{9!}{6!1!2!} = \frac{362880}{720 \cdot 1 \cdot 2} = 252
   \]

Adding all these coefficients together, we get:
\[
126 + 504 + 252 = 882
\]

The final answer is \(\boxed{882}\).