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.
 A: 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.
A: 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.
A: Consider the numbers betweeb $a00$ and $a99$.
Such a number is $ajk$ and $j+k= 10 -a$.  Now $j$ may be anything from $0$ to $10-a$.  That is $11-a$ choices.  $j$ must bue $10-a-j$ and is fixed for us.
(For example:  for then numbers between $600$ and $699$ the number $6jk$ must be that $j+k =4$ and that $j$ may be $0$ to $4$ and there are $5$ such options.  That is $j = 0,1,2,3,4$ and $k= 4,3,2,1,0$.)
So for $a =1......9$ we will have $11-a$ options in that range and so $10+9+ 8+......+2=54$  (the very last case is the numbers between $900$ and $999$ where $9jk$ can have $j = 0,1$ and $k =1,0$.
I think the answer given was careless and accidently included the number $(10)00$ where the first "digit" $a$ is $10$ and $j$ can go from $0$ to $0$.
If you are doing a sum, its easy to think you must go all the way down to $1$.
