(1 + x + x^2)^10

We want the coefficient of $x^7$ in $(1 + x + x^2)^{10} \; = \; \left(\frac{1-x^3}{1-x}\right)^{10} \; = \; (1 - x)^{-10} \big(1 - 10x^3 + 45x^6 + \cdots\big) \; =\; \left(\sum_{n=0}^\infty \binom{n+9}{9} x^n\right)\big(1 - 10x^3 + 45x^6 + \cdots )$ namely $\binom{16}{9} - 10\times\binom{13}{9} + 45\times\binom{10}{9} \; = \; \boxed{4740}$

From each of the 10 factors we have to take one power: or x^0, or x^1, or x^2. Choose 3 times x^2, 1 times x^1 and 6 times x^0. This can be done in $\frac{10!}{3! x 1! x 6!}$ = 840 ways. Or choose 2 times x^2, 3 times x^1 and 5 times x^0 in $\frac{10!}{2! x 3! x 5!}$ = 2520 ways. Or choose 1 times x^2, 5 times x^5 and 4 times x^0 in $\frac{10!}{1! x 5! x 4!}$ = 1260 ways. Or choose 7 times x^1 and 3 times x^0 in $\frac{10!}{7! x 3!}$ = 120 ways. The coefficient of x^7 is 840 + 2520 + 1260 + 120 = 4740.

. We must look of the combination of the coefficient of x^7 from (1)^i . (x)^j . (x^2)^k . where i+j+k = 10 So we have (1)^3 . (x)^7 . (x^2)^0 . there are 10C3 .7C7 . 0C0, and (1)^4 . (x)^5 . (x^2)^1 . there are 10C4 6C5 . 1C1, and (1)^5 . (x)^3 . (x^2)^2 . there are 10C5 5C3 . 2C2,and (1)^6 . (x)^1 . (x^2)^3 . there are 10C6 4C1 . 3C3, total : 10C3 .7C7 . 0C0 + 10C4 6C5 . 1C1 + 10C5 5C3 . 2C2+10C6 4C1 . 3C3 = 4740

This is an expansion of a trinomial

$\sum _{p=0}^3 \frac{10!}{(6-p)! (2 p+1)! (3-p)!} \Rightarrow 4740$ or $\sum _{p=0}^3 \left(\left(\frac{10}{6-p,\,2 p+1,\,3-p}\right)\right) \Rightarrow 4740$

The terms in the denominator, from left to right, respectively are the powers of the $1$ , $x$ and $x^2$ terms. Note that the combined powers of $x$ is 7.