The total number of 3- digit numbers, whose sum of digits is 10 is __________

$\begin{array}{l}\text{ Let }abc\text{ be three digit number }\\ a+b+c=10 where a\ge 1and b,c\ge 0\\ \text{ Let }a-1=x\Rightarrow a=x+1 and x\ge 0\end{array}$

$\begin{array}{l}\therefore x+1+b+c=10\Rightarrow x+b+c=9\\ \text{ Total number of non‐negative solutions }={}^{9+3-1}C_{3-1}={}^{11}C_2=55\end{array}$

But in these solutions (10, 0, 0) is one of the solutions. Eliminate it, the anwer is 54.

Alliter : 

The number of 3 digit numbers, with sum of digits of each number equals to 10 is given by 

the coefficient of $x^{10 } in (x+x^2+x^3+\cdots \cdots +x^9) x {(1+x+x^2+\cdots \cdots +x^9)}^2$

= Coefficient of $x^9 in (1+x+x^2+\cdots +x^8) x (1+x+x^2+\cdots +x^9)$

= Coefficient of  $x^9 in \left(\frac{1-x^9}{1-x}\right) x {\left(\frac{1-x^{10}}{1-x}\right)}^2$

 = Coefficient of $x^9 in \left(1-x^9\right) x {\left(1-x\right)}^{-2}$

= Coefficient of $x^9 in \left(1-x^9\right) x\frac{1 }{2}\left(1.2 + 2.3.x + 3.4.x^2 + \cdots +10.11.x^9 + \cdots \cdots \infty \right)$

= $\frac{1}{2}\left(1.10.11 - 1.1.2 \right) = 55 - 1 =54$