The coefficient of $x^4$in the expansion of ${(1+x+x^2)}^{10}$ is__________

$\text{ General term }=\frac{10!}{\alpha !\beta !\gamma !}\times x^{\beta +2\gamma },\text{ where }\alpha +\beta +\gamma =10$

$\text{ For coefficient of }{x}^4,\beta +2\gamma =4$

So, the possible sets will be

$\begin{array}{l}\gamma =0,\beta =4,\alpha =6 \Rightarrow \frac{10!}{6!4!0!}=210\\ \begin{array}{lll}\gamma =1,\beta =2,\alpha =7& \Rightarrow \frac{10!}{7!2!1!}=360& \\ \gamma =2,\beta =0,\alpha =8\Rightarrow & \frac{10!}{8!0!2!}=45& \\ & & \end{array}\end{array}$

Total = 615