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Equation: $x+y=10$ and we are asked to find out the number of a non-negative integral solution of this equation.
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We just need to find coefficient of $x^{10}$ in the expression $\frac{1}{(1-x)^2}$ which can be written as $\sum_{r=0}^{\infty}\binom{n+r-1}{r}x^r$

                                                                          Here $n=2$    $\Rightarrow\sum_{r=0}^{\infty}\binom{r+1}{r} x^r$

To have co-efficient of $x$ as $10$, we need r to be $10$     $=\binom{10+1}{10}=\binom{11}{10}=11$
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$1 )\ $Brute Force method :


$=\color{Blue}{\text{(X,Y)}}  \ (0,10)  \ (1,9) \ (2,8) \ (3,7) \ (4,6) \ (5,5) \ (6,4) \ (7,3) \ (8,2) \ (9,1) \ (10,0) \\ =\color{Red}{11}  \text{ Pairs}$

$2 )\ $Function method :

  • Looking above values both X and Y can have values in range $(0-10)$.

$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})(1+x+x^{2}+x^{3}....+x^{10})$

$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})^{2}$

$\rightarrow \ [x^{10}]$ $\Large\frac{(1-x^{11})^2}{(1-x)^2}$

$\rightarrow \ [x^{10}]$ $(1+x^{22}-2x^{11})(1-x)^{-2}$

$\rightarrow[x^{10}]$ $\Large{(^{10+2-1}_{ \  \ \  \  \ 10})}$

$\rightarrow$ $\Large(^{11}_{10})$

$\rightarrow\color{Blue}{11.}$

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