Question

The number of points, having both
co-ordinates as intergers, that lie in the interior of the triangle with
vertices $\left(0,0\right),$ $\left(0,41\right)$ and $\left(41,0\right)$ is :

Answer 1

Integral points that line interior are $(x,y)$ when $x>o,y>o$ but $x+y<41$

that means find positive integral solutions of equation $x+y<41$

$x>0,y>0$ so at least $x=1,y=1$

So that means

$x+y\le 39$

So that $x,y$ can take $0..........39$, any value

Total points $=$ (No. of nonintergral solutions of $x+y=0,x+y=1,....x+y=39$).

$={}^1{C}_2+{}^2{C}_1+{}^3{C}_1$

$=1+2+3....40$

$=\frac{40\left(41\right)}{2}=820$.