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  4. The number of non-negative integral solutions of x1+x2+x3+4 x4=20 is

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Q. The number of non-negative integral solutions of $x_{1}+x_{2}+x_{3}+4 x_{4}=20$ is

Permutations and Combinations

Solution:

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 $\left(1+{ }^{3} C_{1} x+{ }^{4} C_{2} x^{2}+{ }^{5} C_{3} x^{3}+{ }^{6} C_{4} x^{4}+\ldots\right.\left.+{ }^{10} C_{8} x^{8} \ldots+{ }^{14} C_{12} x^{12}+\ldots+{ }^{18} C_{16} x^{16}+\ldots+{ }^{22} C_{20} x^{20}+\ldots\right)\times\left(1+x^{4}+x^{8}+x^{12}+x^{16}+x^{20}+\ldots\right)$
$=1+{ }^{6} C_{4}+{ }^{10} C_{8}+{ }^{14} C_{12}+{ }^{18} C_{16}+{ }^{22} C_{20}$
$=1+{ }^{6} C_{2}+{ }^{10} C_{2}+{ }^{14} C_{2}+{ }^{18} C_{2}+{ }^{22} C_{2}$
$=1+\left(\frac{6.5}{1.2}\right)+\left(\frac{10.9}{1.2}\right)+\left(\frac{14.13}{1.2}\right)+\left(\frac{18.17}{1.2}\right)+\left(\frac{22.21}{1.2}\right)$
$=1+15+45+91+153+231=536 .$