Equal to 5

There are 3 arrangements with digits 0,0,5; 6 with 0,1,4; 6 with 0,3,2; 3 with 0,1,3 and 3 with 1,2,2

A total of $\boxed{21}$

The generalized solution is $C_{7,2}=21$ For any number, there are invisible bars separating nearby digits. Accordingly, since we want sum of digits of number,which is between $1$ to $1000$ , to be $5$ ; we therefore have to consider the three-digit numbers, which each has two invisible bars. So, there are $7$ empty boxes at the beginning and two of those to be chosen. Then the answer jumps out! :)

ps, three-digit number in this method includes two-digit and one-digit number.

One solution could to be to transform it into a Combinations-with-repetition problem. Instead of thinking of the different numbers competing for the digits of the number, think of the digits of the number competing for the "ones" that will end up summing to five. For example, 320 could be thought of as: (1-3) (1-3) (1-3) (1-2) (1-2) or even (1-2) (1-3) (1-2) (1-3) (1-3). Because the items (the digit number) are repeating, and their order is irrelevant, this follows the formula for combinations with repetition. The formula is given by (n+r-1)!/r!(n-1)!

One important thing to notice is that, if no "one" is ascribed to the hundreds digit or the tens digit, we will only have a one-digit number. If at least one of the "ones" is given to the tens digit, we will have a two-digit number, while a three-digit number requires at least one "one" for the hundreds digit. Thus, the model of this problem accounts for not only the one-digit numbers, but also the two and three-digit numbers; it can even be extended to numbers with thousands of digits, if necessary.

We have three items (the digit numbers) to fit in five slots (the "ones" that the digits will accumulate to sum up to 5). Our formula is (3+5-1)!/5!(3-1)!, or 7!/5!2!, which evaluates to 21. There are, in fact, 21 different numbers between 1 and 1000 whose digits sum up to 5.