Question

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

• A. 436
• B. 536
• C. 602
• D. None of these

Answer 1

Answer: (B)

Number of non-negative integral solutions of the given equation
$=$ coefficient of $x^{20}$ in
$(1-x{)}^{-1}(1-x{)}^{-1}(1-x{)}^{-1}{\left(1-x^4\right)}^{-1}.$
$=$ coefficient of $x^{20}$ in $(1-x{)}^{-3}{\left(1-x^4\right)}^{-1}$
$=$ coefficient of $x^{20}$ in $(1+{}^3{C}_{1}x+{}^4{C}_2{x}^2+{}^5{C}_3{x}^3+{}^6{C}_4{x}^4+\dots +{}^{10}{C}_8{x}^8\dots +{}^{14}{C}_{12}{x}^{12}+\dots +{}^{18}{C}_{16}{x}^{16}+\dots +{}^{22}{C}_{20}{x}^{20}+\dots )\times \left(1+x^4+x^8+x^{12}+x^{16}+x^{20}+\dots \right)$
$=1+{}^6{C}_4+{}^{10}{C}_8+{}^{14}{C}_{12}+{}^{18}{C}_{16}+{}^{22}{C}_{20}$
$=1+{}^6{C}_2+{}^{10}{C}_2+{}^{14}{C}_2+{}^{18}{C}_2+{}^{22}{C}_2$
$=1+\left(\frac{6.5}{1.2}\right)+\left(\frac{10.9}{1.2}\right)+\left(\frac{14.13}{1.2}\right)+\left(\frac{18.17}{1.2}\right)+\left(\frac{22.21}{1.2}\right)$
$=1+15+45+91+153+231=536.$