The question is from LOD-2 Modern Maths Q22

The question is from LOD-2 Modern Maths Q22

N^(k) _ d means the total possible natural numbers that have at most k digits and the sum of digits is at most d.Now, N^5 _ 5 means that at most 5 digits and the sum is 5. So, the possible cases are:1 digit -> 1, 2, 3, 4, 5 (5 cases)2 digit -> 14, 13, 12, 11, 10, 23, 22, 21, 20, ..., 50 (15 cases)3 digit -> 100 to 104, 110 to 113, 120 to 122, 130 to 131, 140, 200 to 203, 210 to 212 and so on.So using this approach, you can solve the question. If you have any further concerns, let me know.

I am unable to solve this question, I want to understand the approach.

I am unable to solve this question, I want to understand the approach.