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.

Your solution is correct.
Let $h$ denote the hundreds digit, $t$ denote the tens digit, and $u$ denote the units digit.  Then
$$h + t + u = 10 \tag{1}$$
where $1 \leq h \leq 9$, $0 \leq t \leq 9$, and $0 \leq t \leq 9$.
If we let $h' = h - 1$, then $h'$ is a nonnegative integer such that $0 \leq h' \leq 8$.  Substituting $h' + 1$ for $h$ in equation 1 yields
\begin{align*}
h' + 1 + t + u & = 10\\
h' + t + u & = 9 \tag{2}
\end{align*}
which is an equation in the nonnegative integers.
A particular solution of equation 2 corresponds to the placement of $3 - 1 = 2$  addition signs in a row of $9$ ones.  For instance,
$$1 1 1 1 + 1 1 1 + 1 1$$
corresponds to the solution $h' = 4, t = 3, u = 2$ of equation 2 and $h = 5, t = 3, u = 2$ of equation 1.  The number of solutions of equation 2 in the nonnegative integers is
$$\binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} = 55$$
since we must which two of the eleven positions required for nine ones and two addition signs to be filled with addition signs.  This is presumably how the authors of the official solution arrived at the answer $55$.
However, we have not dealt with the restriction that $h' \leq 8$.  There is one solution of equation 2 which violates this restriction, namely $h' = 9, t = 0, u = 0$, which corresponds to the solution $h = 10, t = 0, u = 0$ of equation 1 in the nonnegative integers.  However, if $h = 10$, then we are counting $1000$ as a three-digit integer, which is nonsense.  Subtracting this case, we obtain
$$\binom{9 + 3 - 1}{3 - 1} - 1 = 55 - 1 = 54$$
three-digit positive integers with digit sum $10$, as you found.
