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How many solutions are there to the equation
x1 + x2 + x3 + x4 + x5 = 21,
where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0$\leq$ x1$\leq$10 ?
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$\begin{align*}
&\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 21 \quad \left ( 0\leq x_i \leq 10 \right ) \\ \\
&\Rightarrow \text{No of integral solution of the above equation is } \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left ( 1+x+x^2+ \dots + x^{10} \right )^5 \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ \frac{1-x^{11}}{1-x} \right ]^5 \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ 1-x^{11} \right ]^5 \cdot \left [ \frac{1}{\left ( 1-x \right )^5} \right ] \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \left [ \sum_{m=0}^{5}\binom{5}{m}\left ( -1 \right )^m \cdot x^{11\cdot m} \right ] \cdot \left [ \sum_{r=0}^{\infty}\binom{5+r-1}{r}\cdot x^r \right ] \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} \begin{cases}
&+1 \cdot \binom{25}{21} \qquad m = 0,r = 21 \\ \\
&-5 \cdot \binom{14}{10} \qquad m = 1,r = 10 \\
\end{cases} \\
&\Rightarrow {\color{red}{\left [ x^{21} \right ]}} = +1 \cdot \binom{25}{21} -5 \cdot \binom{14}{10} = {\bf \color{red}{7645}} \\
\end{align*}$


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