Let xyz be the three-digit number such that
x+y+z=10,x≥1,y≥0,z≥0
Let x−1=t⇒x=1+t
So, x≥1⇒t+1≥1⇒t≥0
Also, x+y+z=10⇒t+y+z=10−1=9
So, t+y+z=9,0≤t,y,z≤9
Thus, total number of non-negative integral solution =C3−19+3−1=C211=211.10=55
But for t=9,x=10, so required number of integers =55−1=54.