Number of possible combinations involving the digits 1-9 (inclusive)

Question

Here's the problem (per a Sudoku puzzle).

Suppose I want to know the total number of combinations (order is NOT important) using ANY of the digits from 1-9 (inclusive) and WITHOUT repetition. Furthermore the sum total of the digits and the number of digits in the combination are a Given. The sum of the digits must equal the stated sum

For example: Sum = 10; Number of Digits = 2; Number of combinations = 4 (must be the solved value).

Listing as follows: (9,1); (8,2); (7,3); and (6,4). Two digit combo's are easy to calculate manually, which is what I did here.

Now assume: Sum = 17; Num of Digits = 3; Num of Combo's = 8 (again determined manually - and I might have missed one or two!).

For example:(3,6,8); (3,5,9); (4,6,7); etc.
A number cannot repeat in a grouping, but can repeat in another grouping - as does the digit 3 in the first two listed groups.

So, is there a way to solve for the number of possible combinations with the two Givens and stated condition as indicated??

Someone please say: "Yes, here's how".

joey

Ill have to go through it but it looks good - and very professional too!! — Joey, Mar 17 '21 at 15:16
J. Did you try the formula on my examples. Use the sum of 17 and three digits. If I didn't miss anything, the result should be 8. thanks — Joey, Mar 17 '21 at 15:28
Look at the $x^3q^{17}$ term of the expansion of $\prod\limits_{j=1}^9(1+xq^j)$ and read the coefficient. It should have been $7$ outcomes, not $8$, so you must have made an error. — JMoravitz, Mar 17 '21 at 15:46
Yes!!I counted (3,6,8) and (8,6,3) as two separate combinations — Joey, Mar 17 '21 at 15:51

Number of possible combinations involving the digits 1-9 (inclusive)

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