Let the 3-digit number be $\overline{abc}$, where:

• $a$ is the hundreds digit, so $a \in \{1,2,\dots,9\}$
• $b,c \in \{0,1,2,\dots,9\}$

We need: $a+b+c=10$

Since $a\ge 1$, define: $x=a-1$ Then $x\ge 0$, and the equation becomes: $x+b+c=9$

where $x,b,c \ge 0$.

Now count the number of non-negative integer solutions to: $x+b+c=9$

Using stars and bars, the number of solutions is: $\binom{9+3-1}{3-1}=\binom{11}{2}=55$

Check digit upper bounds:

• $a=x+1 \le 9 \Rightarrow x\le 8$
• $b,c\le 9$

In the equation $x+b+c=9$, $b$ and $c$ cannot exceed $9$ automatically. The only invalid case is when $x=9$, which gives: $x=9,\ b=0,\ c=0$ Then: $a=x+1=10$ which is not a valid digit.

So subtract this one invalid case: $55-1=54$

Therefore, the total number of 3-digit numbers whose digits sum to $10$ is: $\boxed{54}$