Question

Find the coefficient of $x^7$ in the expansion of ${(1+3x-{2x}^3)}^{10}$.

• A. $\frac{10!}{5!4!1!}{\left(1\right)}^5{\left(3\right)}^4{(-2)}^1+\frac{10!}{7!2!1!}{\left(1\right)}^7{\left(3\right)}^1{(-2)}^2$
• B. $\frac{10!}{3!7!0!}{\left(1\right)}^3{\left(3\right)}^7{(-2)}^0+\frac{10!}{5!4!1!}{\left(1\right)}^5{\left(3\right)}^4{(-2)}^1$
• C. $\frac{10!}{3!0!}{\left(1\right)}^3{\left(3\right)}^7{(-2)}^0+\frac{10!}{5!4!}{\left(1\right)}^5{\left(3\right)}^4{(-2)}^1+\frac{10!}{7!1!}{\left(1\right)}^7{\left(3\right)}^1{(-2)}^2$
• D. $\frac{10!}{3!7!0!}{\left(1\right)}^3{\left(3\right)}^7{(-2)}^0+\frac{10!}{5!4!1!}{\left(1\right)}^5{\left(3\right)}^4{(-2)}^1+\frac{10!}{7!2!1!}{\left(1\right)}^7{\left(3\right)}^1{(-2)}^2$

Answer 1

Answer: (D)

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

The formula to find the coefficient of $a^p{b}^q{c}^r{d}^s$ in $(xa+yb+zc+vd{)}^n$ is $\frac{n!x^p{y}^q{z}^r{v}^s}{p!q!r!s!}$

$x^7$ will be possible in 3 ways.
$(-2x^3{)}^2.(3x\left)\right(1^7)$
coefficients is $(-2{)}^2(3\left)\right(1)\frac{10!}{7!.2!.1!}$

$(-2x^3)\left(3x{)}^4\right(1{)}^5$
coefficient is $(-2)\left(3^4\right)(1{)}^5\frac{10!}{4!.1!.5!}$

$\left(2x^3{)}^0\right(3x{)}^7(1{)}^3$
Coefficient is $(3{)}^7{1}^3\frac{10!}{7!.3!}$