$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 $(0-10)$.
$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})(1+x+x^{2}+x^{3}....+x^{10})$
$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})^{2}$
$\rightarrow \ [x^{10}]$ $\Large\frac{(1-x^{11})^2}{(1-x)^2}$
$\rightarrow \ [x^{10}]$ $(1+x^{22}-2x^{11})(1-x)^{-2}$
$\rightarrow[x^{10}]$ $\Large{(^{10+2-1}_{ \ \ \ \ \ 10})}$
$\rightarrow$ $\Large(^{11}_{10})$
$\rightarrow\color{Blue}{11.}$