Given a n number of digits and a sum x, how can I count the amount of ways you can obtain x from the sum of the digits?

Asked Mar 22 '23 at 11:29

Active Mar 22 '23 at 12:09

Viewed 29 times

My problem originally was "Given a number of two digits, in how many ways can you obtain 12 as the sum of its digits?".

I could find the solution by counting:

93: 9 + 3
84: 8 + 4
75: 7 + 5
66: 6 + 6
57: 5 + 7
48: 4 + 8
39: 3 + 9

Then I found 7 as the answer. But I realized that if the number of digits were 3, the counting would get too big already, so I would like to know if there's a way to generalize this problem and its solution.

edited Mar 22 '23 at 12:09

asked Mar 22 '23 at 11:29

Jonas

1

Just to be clear, in your mind, digits go from 0 to 9? So when $n=3$ and $x = 1$, you can write $x = 0 + 0 + 1$? And when $n = 3$ and $x=28$, there is no solution because $9+9+9<28$? – cs89 Mar 22 '23 at 11:35
yes! digits go from 0 to 9 – Jonas Mar 22 '23 at 11:37
1

But can you have a number 007 ? – true blue anil Mar 22 '23 at 11:39
@trueblueanil yes, if n = 3. Also 070 and 700 – Jonas Mar 22 '23 at 12:10

Given a n number of digits and a sum x, how can I count the amount of ways you can obtain x from the sum of the digits?

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