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

We have to find the coefficient of $$x^4$$ in the expansion of $$(1 + x + x^2)^{10}$$.

The expression $$(1 + x + x^2)^{10}$$ has three terms being raised to the 10th power. The appropriate tool is the multinomial theorem, which states:

For $$(a + b + c)^{n}$$, a general term is $$\dfrac{n!}{n_1!\,n_2!\,n_3!}\;a^{\,n_1} b^{\,n_2} c^{\,n_3}$$ where $$n_1 + n_2 + n_3 = n$$.

Here we identify $$a = 1,\; b = x,\; c = x^2,\; n = 10$$. After substitution, a general term becomes

$$\dfrac{10!}{n_1!\,n_2!\,n_3!}\;1^{\,n_1} x^{\,n_2} (x^2)^{\,n_3} \;=\; \dfrac{10!}{n_1!\,n_2!\,n_3!}\;x^{\,n_2 + 2n_3}.$$

We need the power of $$x$$ to be exactly 4, so we set

$$n_2 + 2n_3 = 4.$$

Because $$n_1 + n_2 + n_3 = 10$$ and all $$n_i$$ are non-negative integers, we solve $$n_2 + 2n_3 = 4$$ for possible pairs $$(n_2, n_3)$$:

• If $$n_3 = 0$$ then $$n_2 = 4$$, giving $$n_1 = 10 - 4 - 0 = 6.$$br • If $$n_3 = 1$$ then $$n_2 = 2$$, giving $$n_1 = 10 - 2 - 1 = 7.$$br • If $$n_3 = 2$$ then $$n_2 = 0$$, giving $$n_1 = 10 - 0 - 2 = 8.$$

No other non-negative solutions exist, so there are three contributing terms.

Now we calculate the coefficient for each set $$(n_1, n_2, n_3)$$ using the multinomial coefficient $$\dfrac{10!}{n_1!\,n_2!\,n_3!}$$.

For $$(6,4,0):\; \dfrac{10!}{6!\,4!\,0!} =\dfrac{10\times9\times8\times7}{4\times3\times2\times1} =210.$$

For $$(7,2,1):\; \dfrac{10!}{7!\,2!\,1!} =\dfrac{10\times9\times8}{2\times1} =360.$$

For $$(8,0,2):\; \dfrac{10!}{8!\,0!\,2!} =\dfrac{10\times9}{2\times1} =45.$$

Adding all these contributions:

$$210 + 360 + 45 = 615.$$

So, the answer is $$615$$.