# Amount of numbers with certain condition

I'm sorry I have asked until now up to three questions regarding the same topic, but tomorrow I have a Combinatorics exam and I'm practising in order to be ready and do my best. Here is the question: I am requiered to find how many numbers in the set $$\{1\dots 1000\}$$ there are whose sum of digits is $$5$$. Here is my attempt:
I started separating in cases, depending on how many digits there were:

1. If we only have one digit, it is easy: only the number $$5$$ satisfies it: 1 option
2. If we have two digits, it is the same as solving $$x_1 + x_2 = 5$$, where $$x_i$$ is the $$i$$-digit and $$x_1 \geq 1, x_2 \geq 0$$. The number of solutions is, if I'm not wrong, $${5 \choose 1} = 5$$
3. If we have three digits, the argument is the same as above, but with three variables: $$x_1 + x_2 + x_3 = 5, x_1 \geq 1, x_2,x_3 \geq 0$$. The number of solutions is $${6 \choose 2} = 15$$. Therefore, and since cases are mutually exclusive, the result would be the sum of these three options: $$21$$. Am I right?
• Looks good. $\quad$
– lulu
Commented Oct 23, 2023 at 19:09
• As a mechanical check, the direct count is pretty easy. The only challenge is for the three digit numbers, but you only need to look at the numbers in $\{104, 113, \cdots, 500\}$ and that's not that bad.
– lulu
Commented Oct 23, 2023 at 19:12

Your argument works, but there's a more elegant solution!

Allow leading zeroes. Ignoring the number $$1000$$, whose digits don't sum to $$5$$ anyway, treat each number as a $$3$$-digit string $$d_2d_1d_0$$. So we're looking to enumerate integer solutions to $$d_0 + d_1 + d_2 = 5,$$ where each $$d_i \geq 0$$. This is a stars-and-bars calculation: $$\binom{5+3-1}{3-1} = \binom{7}{2} = 21.$$

• I still cannot understand how I couldn't come up directly with that. Thanks for your answer Commented Oct 23, 2023 at 19:19