How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$, 
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6$ in $(x+(1+i))^{10}(x+(1-i))^{10}$, then dividing by $2^{10}$.
Then, we find that the coefficient of $x^6$ would be:

$$\sum_{i=0}^{6} \binom{10}{6-i} \binom{10}{i} (1-i)^{10-(6-i)} (1+i)^{10-i}$$

with the knowledge that $(1-i)(1+i)=2$, I simplified to

$$\binom{10}{6}\binom{10}{0}2^4((1-i)^6+(1+i)^6)+\binom{10}{5}\binom{10}{1}2^5((1-i)^4+(1+i)^4)+\binom{10}{4}\binom{10}{2}2^6((1-i)^2+(1+i)^2)+\binom{10}{3}\binom{10}{3}2^7$$

Note: the formula $(1+i)^x+(1-i)^x$ gives:
$ 2(2^{\frac{x}{2}}) \cos(\frac{x\pi}{4})$
After simplifying and reapplying the division by $2^{10}$, I get $(\frac{0}{1024}) + (-8)(2520)(\frac{32}{1024}) + (\frac{0}{1024}) + (120)(120)(2^7)$, which gives $0-630+0+1800,$ which is 1170, and I checked this over with an expression expansion calculator.
If the original equation was $(1+x+x^2)^{10}$, I would have used binomials to find the answer, however, the $x^2$ was replaced by $\frac{x^2}{2}$. 
My question is whether anyone has a combinatorics solution to this question, rather than just algebra. It would be nice if the solution did not require complex numbers.

Write the expression as $\frac{1}{2^{10}}\left((x+1)^2+1\right)^{10}$. 
Hint if you would like to do it using combinatorial arguments : Coefficient of $x^6$ in $(1+x^2)^{n}$ would give you the number of tuples such that $a_1 + a_2 + \cdots a_{n} = 6$ such that $a_i\in \{0,2\}$. This should be easy- just note that any three of the $a_i$s should be $2$ which gives $n\choose 2$ as the answer to the example here.
To add to the hint, note that if you write $x+1=t$, then you need to calculate the coefficients of $x^6$ which you get directly from the coeff of $t^6$ and another from the coeff of $t^4$ because $t^4 = (1+x)^4.
