Triangle is formed by the co-ordinates $(0, O), (0, 21)$ and $(21, O)$.Then the numbers of integral co-ordinate strictly inside the triangle (integral co-ordinates of both x and y) 

The three points form a right angled triangle whose hypotenuse by intercepts form is $\frac{x}{21}+\frac{y}{21}=1$ or

$x + y = 21$. We have to find the number of points having integral coordinates and lying within the triangle. In other words we have to find the number of points having integral solution of the inequality $x + y < 21$or

$x<21-y,\text{ where }0<x,y<21$

For $x = 1, y = 20$ but the point (1, 20) lies 011 the line AB and hence excluded. So for x = 1 we can have $y = 1, 2, ... , 19$ (20 i" excluded}. Similarly, for $x = 2$, we can have $y = 1, 2, ... , 18$ and so on. Hence the total is

$19+18+17+\dots +1=\mathrm{\Sigma }n=\frac{n(n+1)}{2}=\frac{19\cdot 20}{2}=190.$