Question

The coefficient of $x^{10}$ in the expansion of $[1+x^2(1-x){]}^8$ is

• A. $470$
• B. $400$
• C. $576$
• D. $476$

Answer 1

The correct option is D $476$

We have,
$[1+x^2(1-x){]}^8=(1+x^2-x^3{)}^8=\sum _{k_1+k_2+k_3=8}\frac{8!}{k_1!k_2!k_3!}\left(1{)}^{k_1}\right(x^2{)}^{k_2}(-x^3{)}^{k_3}=\sum _{k_1+k_2+k_3=8}\frac{8!}{k_1!k_2!k_3!}(-1{)}^{k_3}{x}^{2k_2+3k_3}$

Coefficient of $x^{10}$, so
$2k_2+3k_3=10k_1+k_2+k_3=8$

So,
$\left(1\right)k_1=4,k_2=2,k_3=2\left(2\right)k_1=3,k_2=5,k_3=0$

Therefore, the required coefficient
$=\frac{8!}{4!2!2!}+\frac{8!}{3!5!0!}=\frac{8\cdot 7\cdot 6\cdot 5}{4}+\frac{8\cdot 7\cdot 6}{6}=420+56=476$