1. We need the coefficient of $x^7$ in

$$(1-x+2x^3)^{10}.$$

2. In each factor, we choose one of $1$, $-x$, or $2x^3$.

Suppose:

• $a$ factors contribute $1$
• $b$ factors contribute $-x$
• $c$ factors contribute $2x^3$

Then:

$$a+b+c=10$$

and the power of $x$ contributed is

$$b+3c.$$

We want

$$b+3c=7.$$

3. Solve for non-negative integer solutions of

$$b+3c=7.$$

Possible values:

• If $c=0$, then $b=7$.
• If $c=1$, then $b=4$.
• If $c=2$, then $b=1$.

So the valid triples $(a,b,c)$ are:

• $(3,7,0)$
• $(5,4,1)$
• $(7,1,2)$

4. Compute contribution from each case.

For fixed $(a,b,c)$, coefficient contribution is

$$\frac{10!}{a!b!c!}(1)^a(-1)^b(2)^c.$$

Case 1: $(a,b,c)=(3,7,0)$

$$\frac{10!}{3!7!}(-1)^7 = \binom{10}{7}(-1) = 120(-1)=-120.$$

Case 2: $(a,b,c)=(5,4,1)$

$$\frac{10!}{5!4!1!}(1)^5(-1)^4 2^1 = \frac{10!}{5!4!} \cdot 2.$$

Now,

$$\frac{10!}{5!4!1!}=1260.$$

Hence contribution is

$$1260\cdot 2=2520.$$

Case 3: $(a,b,c)=(7,1,2)$

$$\frac{10!}{7!1!2!}(1)^7(-1)^1 2^2.$$

Now,

$$\frac{10!}{7!1!2!}=360,$$

so contribution is

$$360\cdot(-1)\cdot 4=-1440.$$

5. Add all contributions:

$$-120+2520-1440=960.$$

Therefore, the coefficient of $x^7$ is

$$\boxed{960}.$$

6. Comparison with stored correct answer:

• Derived answer: $960$
• Stored correct answer: $960$

They match.