Question

The coefficients of $x^2,y^5,z^3$ in the expansion of ${(2x+y+3z)}^{10}$ is $k.10!\times 3^3$ then $k$ equals

• A. $\frac{10!}{2!5!3!}{2}^2{3}^2$
• B. $\frac{10!}{5!3!}{3}^3{2}^2$
• C. $\frac{1}{3.5!}$
• D. $\frac{10!}{2!5!3!}{2}^2{3}^3$

Answer 1

The correct option is C $\frac{1}{3.5!}$

${(2x+y+3z)}^{10}$ $=\sum _{x_1+x_2+x_3=10}\frac{101}{x_1!x_2!x_3!}{\left(2x\right)}^{x_1}{\left(y\right)}^{x_2}{\left(3z\right)}^{z_3}$ General term in the expression is $\frac{10!}{x_1!x_2!x_3!}{\left(2x\right)}^{x_1}{\left(y\right)}^{x_2}{\left(3z\right)}^{x_3}=\frac{10!}{x_1!x_2!x_3!}{2}^{x_1}{3}^{x_3}{x}^{x_1}.y^{x_2}{z}^{x_3}$

Now for the coefficient of $x^2{y}^5{z}^3$ we have $x_1=2,x_2=5,x_3=3$ $\therefore$ Required coefficient of $x^2{y}^5{z}^3$ is

$\frac{10!}{2!5!3!}{2}^2{3}^3$

Now $\frac{10!}{2!5!3!}{2}^2{3}^3=k10!\times 3^3$

$\therefore k=\frac{4}{2!3!5!}=\frac{4}{\mathrm{2.6.120}}=\frac{1}{360}=\frac{1}{3.5!}$