How many $3$-digit numbers are there whose sum of digits is $10$? I did just counted all in gap of every $100$ numbers. Like in $100-200$, there are $10$ nos.($109,118,127, \ldots,181,190$). And then $9$ nos. in $200-300$, $8$ in $300-400$ ... and so on. By this I am getting $10+9+...3+2=54$, but answer given in India's JEE Mains exam is $55$. Am I missing one or the answer given is wrong and I'm right? Please help.

This problem is equivalent to partitioning ten items to three boxes.
For example $433$ whose sum of digits is 10. $433$ can be represented as partition of items as below:
$$\circ \circ \circ \circ \mid \circ \circ \circ \mid  \circ   \circ  \circ$$
More examples:
$109$
$$\circ \mid \mid \circ \circ \circ \circ \circ \circ \circ   \circ  \circ$$
$019$, this is not a 3-digit number.
$$\mid \circ \mid \circ \circ \circ \circ \circ \circ \circ   \circ  \circ$$
$190$
$$ \circ \mid \circ \circ \circ \circ \circ \circ \circ   \circ  \circ \mid$$
Now we calculate the Number of three digit number with following three mutually exclusive cases.
Case I: Zero is not one of the digits.
Here we need to choose partition from the $9$ spaces between the items.
This is $= \binom{9}{2} = 36$
Case II: One of the digit except the last digit is zero. (Eg: 109)
Both the partitions occupy the same position.
Number of choices = $\binom{9}{1} = 9$
Case III: The last digit is zero but other digits are non-zero. (Eg: 190)
One of the partitions is at the end and other partition can be chosen from one of the 9 places. The number of choices = $\binom {9}{1} = 9$
Therefore total number of 3-digit numbers whose sum of digits is $10 = 36 + 9 + 9 = 54$.
NOTE: $100, 200, \cdots, 900$ digits with two zeros are not included in any of these cases. All these numbers have a digit sum less than 10.
