$\begin{align*} &\text{We need no of solutions of the following Eq.} \\ & x_1 +x_2 +x_3 = 10 \\ \\ &\text{Where } 0\leq x_1\leq 10 \text{ , }0\leq x_2\leq 5 \text{ , }0\leq x_3\leq 3 \\ \\ \hline \\\\ &\text{Using generating function} \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left ( 1+x+x^2+ \dots + x^{10} \right )*\left ( 1+x+x^2+ \dots + x^{5} \right )*\left ( 1+x+x^2+ x^{3} \right ) \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \frac{1-x^{11}}{1-x} \right ]*\left [ \frac{1-x^{6}}{1-x} \right ]*\left [ \frac{1-x^{4}}{1-x} \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \left ( 1-x^{11} \right )*\left ( 1-x^{6} \right )*\left ( 1-x^{4} \right ) \right ]*\left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \left ( 1-{\color{red}{\bf x^{11}}} \right )*\left ( 1-x^4-x^6+x^{10} \right ) \right ]*\left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ \\ &\text{drop }x^{11} \text{ term} \\ \\ \end{align*}$ $\begin{align*} \\ \\ &\text{Now ,} \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [1-x^4-x^6+x^{10} \right ]*\left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\begin{cases} & {\color{blue}{\bf +}}\;1*\binom{12}{10} = +66 \;\;\;\;\; {\color{blue}{+x^0 \text{ and k=10}}} \\ & {\color{blue}{\bf -}}\;1*\binom{8}{6} \;= -28 \;\;\;\;\; {\color{blue}{-x^4 \text{ and k=6}}} \\ & {\color{blue}{\bf -}}\;1*\binom{6}{4} \;= -15 \;\;\;\;\; {\color{blue}{-x^6 \text{ and k=4}}} \\ & {\color{blue}{\bf +}}\;1*\binom{2}{0} \;= +1 \;\;\;\;\;\; {\color{blue}{+x^{10} \text{ and k=0}}} \\ \end{cases} \\ \\ &\text{Answer } = 66-28-15+1 = 24 \\ \end{align*}$
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