Possible numbers with each digit value range given - combinations

Question

We are given 3 variables, d, a and b.
d specifies the maximum number of places.
a and b specify the minimum digit and maximum digit respectively.
We have to find how many numbers can be made each of which has its digits in non-decreasing order, with maximum d places, and each digit between a and b, both inclusive.
Ex-
d = 1
a = 8
b = 9
Answer = 2 (Possible numbers are 8 and 9)
Ex-
d = 2
a = 8
b = 9
Answer = 5 (Possible numbers are 8, 9, 88, 89, and 99)
I have tried applying, permutation and combinations, but could not come across a generic answer for all values of d.
Also, I tried to observe patterns in possible numbers of each digit, but it seems to be varying.
What is the actual approach for it?

Answer 1

You are looking for combinations with repetition. All we have to do is generate all combinations with repetition of your set for each specific length up to d.
Below, we have a simple implementation in C++. First we need the appropriate include along with a function for return the number of combinations with repetition (i.e. nChooseK):
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
int nChooseK(int n, int k) {
int nCk = 1;
for (int i = (n - k + 1), d = 1; d <= k; ++i, ++d) {
nCk *= i;
nCk /= d;
}
return round(nCk);
}
And here is the algorithm for generating combinations with repetition:
void combsWithRep(std::vector<int> v, int d) {
int numIter, maxZ, count = 0;
int n = v.size();
int d1 = d - 1, d2 = d - 2;
std::vector<int> z(d, 0);
int numRows = nChooseK(n + d - 1, d);
maxZ = n - 1;
while (count < numRows) {
numIter = n - z[d1];;
for (int i = 0; i < numIter; ++i, ++count, ++z[d1]) {
for (int k = 0; k < d; ++k) {
std::cout << v[z[k]];
}
std::cout << std::endl;
}
for (int i = d2; i >= 0; i--) {
if (z[i] != maxZ) {
++z[i];
for (int k = (i + 1); k < d; ++k)
z[k] = z[k - 1];
break;
}
}
}
}
And here is the main function:
int main() {
std::vector<int> v;
int d, input;
std::cout << "Enter the number of possible digits" << std::endl;
std::cin >> d;
std::cout << "\nEnter the digits" << std::endl;
while (std::cin >> input)
v.push_back(input);
std::sort(v.begin(), v.end());
for (int i = 1; i <= d; ++i)
combsWithRep(v, i);
return 0;
}
So, for an input of:
Enter the number of possible digits
4
Enter the digits
7
8
9
The output is:
7
8
9
77
78
79
88
89
99
777
778
779
788
789
799
888
889
899
999
7777
7778
7779
7788
7789
7799
7888
7889
7899
7999
8888
8889
8899
8999
9999