Question

The number of non negative integral solution of $x_1+x_2+x_3+4x_4=20$ is

• A. $536$
• B. $443$
• C. $655$
• D. none of these

Answer 1

The correct option is A $536$

The number of non negative integral solution of given equation is equal to the coefficient of $x^{20}$ in $(1+x+x^2\cdots )(1+x+x^2\cdots )(1+x+x^2\cdots )(1+x^4+x^8\cdots )$
$=\text{coeff. of}{x}^{20}\text{in}(1-x{)}^{-1}(1-x{)}^{-1}(1-x{)}^{-1}(1-x^4{)}^{-1}$
$=\text{coeff. of}{x}^{20}\text{in}(1-x{)}^{-3}\times (1-x^4{)}^{-1}$
$=\text{coeff. of}{x}^{20}\text{in}(1+x^4+x^8+\cdots )\times (1+{}^3{C}_{1}x+{}^4{C}_2{x}^2+{}^5{C}_3{x}^3+{}^6{C}_4{x}^4+\cdots )$
$=1+{}^6{C}_4+{}^{10}{C}_8+{}^{14}{C}_{12}+{}^{18}{C}_{16}+{}^{22}{C}_{20}$
$=536$

Alternate Solutions:
$x_1+x_2+x_3+4x_4=20$
Now puting $x_4=0,1,2,3,4,5$ we get
$x_1+x_2+x_3=20,x_1+x_2+x_3=16\cdots ,x_1+x_2+x_3=4,x_1+x_2+x_3=0$
Now for number of solutions of all the eqaution
$={}^{22}{C}_2+{}^{18}{C}_2+{}^{14}{C}_2+{}^{10}{C}_2+{}^6{C}_2+{}^2{C}_2=231+153+91+45+15+1=536$