# repeated combinations star bars sum digits

I have been looking at some simpler versions where digits sum < 9 example: from 1 to $10^4$ how many sum to 5? using repeated combinations $$\binom{5+5−1}{5}=\binom{9}{5}=126$$ however the given answer is 56

does a generalized version (formula) exist where R = sum of digits ; N = number of digits. $$A \leq N \leq B$$

in the example above$$R = 5; N = 5; A = 1; B = 5;$$

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Your error is that the numbers between $1$ and $10^4$ don't have 5 digits; they have at most $4$ digits, except for the largest one, $10^4 = 10000$, whose digits don't add up to $5$.
Instead, you need to count the number of $4$-tuples that add up to $5$; these will correspond to numbers between $0$ ($=0000$) and $9999$ whose digits add up to $5$. So the correct formula would be $$\binom{4+5-1}{5} = \binom{8}{5} = 56.$$ Your binomial coefficient counts, instead, the number between $1$ and $10^5$ whose digits add up to $5$ (by counting those between $0$ and $99,999$).