If $\alpha =x_1,x_2,x_3\text{ and }\beta =y_1,y_2,y_3$  be two three digits numbers , the number of pairs of $\alpha \text{ and }\beta$ can
be formed so that $\alpha$ can be subtracted from $\beta$ without borrowing is K then K is  equal to 

$\therefore \alpha$  can be subtracted from $\beta$ without borrowing if $y_i\ge x_i;\text{ for }i=1,2,3$ $\text{ Let }{x}_i=\lambda$ 

$\text{ If }i=1\text{, then }\lambda =1,2,3,\dots .,9$ 

$\text{ And for }i=2\text{ and }3,\lambda =0,1,2,3,\dots .,9$

Hence, total number of ways of choosing the pair

$\alpha ,\beta \text{ is }\left(\sum _{\lambda =1}^9 (10-\lambda )\right){\left(\sum ^9 (10-\lambda )\right)}^2\text{=}\left(45\right)(55{)}^2$