How many numbers between $100$ and $900$ have sum of their digits equal to $15$? 
How many numbers between $100$ and $900$ have sum of their digits
  equal to $15$?

I was wondering if we could derive any direct combinatoric formula to solve problem like this,my strategy is  kind of trying all permutation for $100$ to $200$ then $200$ to $300$ then ...,and finally summing them up to get $62$ (if I there is not mistake)though this may not be very difficult to solve using this approach (as we are dealing with $3$ digits numbers with individual digits not greater than $9$)but is this also the fastest/efficient one?
 A: HINT:
$x_1 + x_2 + x_3 = 15, \; x_1 \geq 1, \; x_1, x_2, x_3 < 10$ is very relevant here.
A: This problem can be reduced to solving
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9, \quad x_1 \neq 0, \quad x_1 \neq 9$$
First, we solve
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 15$$
This is a standard stars and bars problem, with solution $\binom{15+3-1}{15} = 136$. Now which solutions are wrong? Well, first of all those with one of them above $10$. But if, say, $x_1 \geq 10$, then we can write $x_1' = x_1 - 10$ and $x_2' = x_2, x_3' = x_3$ to reduce the problem to
$$x_1' + x_2' + x_3' = 5, \quad 0 \leq x_i' \leq 5$$
This is again a standard stars and bars problem, with solution $\binom{5+3-1}{5} = 21$. Since exactly one of $x_1,x_2,x_3$ could be above $10$, this gives $3 \cdot 21 = 63$ wrong solutions in total. So there are $136 - 63 = 73$ solutions to
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9$$
Finally, we need to subtract $4$ solutions with $x_1 = 0$ (i.e. $(x_2,x_3) \in {(6,9),(7,8),(8,7),(9,6)}$), but also subtract $7$ solutions for $x_1 = 9$ (i.e. $(x_2,x_3) \in {(0,6),\ldots,(6,0)}$). This gives $62$ solutions in total.
A: This is almost a standard stars-and-bars (or marbles and boxes) problem. You have $15$ marbles, and you want to distribute them amongst $3$ boxes (the digits) in such a way that the first box gets at least one marble and at most $8$ marbles and each of the other two boxes gets at most $9$ marbles. ($900$ is the only allowable $3$-digit number beginning with $9$, and its digits don’t sum to $15$, so it can be ignored.) Equivalently, you’re looking for non-negative integer solutions to $x_1+x_2+x_3=15$ subject to the constraints $1\le x_1 \le 8, x_2,x_3 \le 9$.
After putting one marble in the first box, you’ve $14$ left to distribute. Ignoring the upper bounds, this can be done in $\dbinom{14+3-1}{3-1}=\dbinom{16}{2}=120$ ways. The ways that have at least $9$ marbles in the first box are bad, so they have to be subtracted off. After putting $9$ marbles in the first box, you’ve $6$ left to distribute; this can be done in $\dbinom{6+3-1}{3-1}=\dbinom{8}{2}=28$ ways. Similarly, there are $\dbinom{4+3-1}{3-1}=\dbinom{6}{2}=15$ distributions that have too many marbles in the second box (because after putting one in the first box and $10$ in the second box you’ve only $4$ left to distribute) and $15$ that have too many in the third box. It’s not possible to have too many marbles in two boxes: that would require at least $19$ marbles. Thus, the final count is $120-28-2(15)=62$ numbers between $100$ and $900$ whose digits sum to $15$.
A: There is a high mass:volume ratio, so here we go ... 
I'll just get you started counting, since other answers are being posted.
$15 = 9 + 6 + 0$. How many permutations of this are between $100$ and $900$?

$15 = 9 + 5 + 1$. How many ...?
$15 = 9 + 4 + 2$...
...

$15 = 8 + 7 + 0$...
$15 = 7 + 7 + 1$...
