The coefficient of x^4 in the expansion of (1+x+x^2^6 is

Choose from each of the six factors either $1$, $x$ or $x^2$. If $j$ factors contribute $x$ and $m$ factors contribute $x^2$, the exponent is $j+2m=4$. The possibilities are $(j,m)=(4,0),(2,1),(0,2)$. Count the arrangements with multinomial coefficients: $(4,0)$ gives $\binom64=15$; $(2,1)$ gives $\dfrac{6!}{2!\,1!\,3!}=60$; $(0,2)$ gives $\binom62=15$. Adding, the coefficient is $15+60+15=90$. Missing the middle case $(2,1)$ gives the wrong total $30$.