The coefficient of x^7 in (1-x+2x^3)^10 is .

We need to find the coefficient of ${x}^{7}$x7 in ${(1-x+2{x}^{3})}^{10}$(1−x+2x3)10.

We know that for ${(x+y+z)}^{n}$(x+y+z)n,

${T}_{n}=\frac{n!}{a!b!c!}{(x)}^{a}{(y)}^{b}{(z)}^{c}$Tn​=a!b!c!n!​(x)a(y)b(z)c such that $a+b+c=n$a+b+c=n

Now for ${(1-x+2{x}^{3})}^{10}$(1−x+2x3)10

$${T}_{n}=\frac{10!}{a!b!c!}{(1)}^{a}{(-x)}^{b}{(2{x}^{3})}^{c}$$Tn​=a!b!c!10!​(1)a(−x)b(2x3)c

$=\frac{10!}{a!b!c!}{(-1)}^{b}{2}^{c}{x}^{b+3c}$=a!b!c!10!​(−1)b2cxb+3c. with $a+b+c=10$a+b+c=10

Here we need coefficient of ${x}^{7}$x7.

Hence the combinations would be

$$\begin{matrix}a & b & c \\ 3 & 7 & 0 \\ 5 & 4 & 1 \\ 7 & 1 & 2\end{matrix}$$a357​b741​c012​

which satisfies both $a+b+c=10$a+b+c=10 and $b+3c=7$b+3c=7.

Hence, coefficient of ${x}^{7}$x7 will be,

$$=\frac{10!}{2!1!7!}{(2)}^{2}{(-1)}^{1}$$=2!1!7!10!​(2)2(−1)1

$$+\frac{10!}{1!4!5!}{(2)}^{1}{(-1)}^{4}+\frac{10!}{0!7!3!}{(2)}^{0}{(-1)}^{7}$$+1!4!5!10!​(2)1(−1)4+0!7!3!10!​(2)0(−1)7

$$=960$$=960