Mathematics - Permutation & Combination Question with Solution | TestHub

Let $xyz$ be the three-digit number such that

$x+y+z=10, x\ge 1, y\ge 0, z\ge 0$

Let $x-1=t\Rightarrow x=1+t$

So, $x\ge 1\Rightarrow t+1\ge 1\Rightarrow t\ge 0$

Also, $x+y+z=10\Rightarrow t+y+z=10-1=9$

So, $t+y+z=9, 0\le t, y, z\le 9$

Thus, total number of non-negative integral solution $={}^{9+3-1}C_{3-1}={}^{11}C_2=\frac{11.10}{2}=55$

But for $t=9, x=10$, so required number of integers $=55-1=54$.