Multinomial Theorem

Using the multinomial theorem:
$$(1 + x + x^2)^{10} = \sum_{i+j+k=10} \binom{10}{i,j,k} 1^i \cdot x^j \cdot (x^2)^k = \sum_{i+j+k=10} \binom{10}{i,j,k} x^{j+2k}$$

For the coefficient of $$x^5$$, we need: $$j + 2k = 5$$ and $$i + j + k = 10$$

From the first equation: $$j = 5 - 2k$$
From the second equation: $$i = 10 - j - k = 10 - (5-2k) - k = 5 + k$$

For non-negative integers:
- $$j = 5 - 2k \geq 0 \Rightarrow k \leq 2.5 \Rightarrow k \leq 2$$
- $$k \geq 0$$

So $$k \in \{0, 1, 2\}$$

For $$k = 0$$: $$i = 5, j = 5, k = 0$$
$$\binom{10}{5,5,0} = \frac{10!}{5! \cdot 5! \cdot 0!} = \binom{10}{5} = 252$$

For $$k = 1$$: $$i = 6, j = 3, k = 1$$
$$\binom{10}{6,3,1} = \frac{10!}{6! \cdot 3! \cdot 1!} = 840$$

For $$k = 2$$: $$i = 7, j = 1, k = 2$$
$$\binom{10}{7,1,2} = \frac{10!}{7! \cdot 1! \cdot 2!} = 360$$

Therefore the final coefficient of $$x^5$$ in the expansion of $$(1 + x + x^2)^{10}$$ $$= 360 + 840 + 252=1452$$