Program to count n digit integers where digits are strictly increasing in Python

Suppose we have a number n, we have to find the number of n-digit positive integers such that the digits are in strictly increasing order.

So, if the input is like n = 3, then the output will be 84, as numbers are 123, 124, 125, ..., 678, 789.

Understanding the Problem

For n-digit numbers with strictly increasing digits, we need to choose n different digits from {1, 2, 3, ..., 9}. We cannot use 0 as the first digit in an n-digit number. This becomes a combination problem: C(9, n).

Algorithm

To solve this, we will follow these steps −

• if n ? 9 and n > 0, then

  • return Combination C(9, n)

• otherwise,

  • return 0

Implementation Using math.factorial

Let us see the following implementation to get better understanding −

from math import factorial

class Solution:
    def solve(self, n):
        if 0 < n <= 9:
            return factorial(9) // factorial(n) // factorial(9 - n)
        else:
            return 0

ob = Solution()
print(ob.solve(3))

The output of the above code is −

84

Using math.comb Function

Python 3.8+ provides a built-in combination function that's more readable −

from math import comb

class Solution:
    def solve(self, n):
        if 0 < n <= 9:
            return comb(9, n)
        else:
            return 0

ob = Solution()
print(f"n = 3: {ob.solve(3)}")
print(f"n = 4: {ob.solve(4)}")
print(f"n = 9: {ob.solve(9)}")
print(f"n = 10: {ob.solve(10)}")

The output of the above code is −

n = 3: 84
n = 4: 126
n = 9: 1
n = 10: 0

How It Works

For n = 3, we choose 3 digits from {1, 2, 3, 4, 5, 6, 7, 8, 9}. The combinations are C(9, 3) = 84. Some examples include 123, 145, 678, 789, etc. Since digits must be strictly increasing, there's only one way to arrange the chosen digits.

Conclusion

The problem reduces to calculating combinations C(9, n) since we select n digits from 9 available digits (1-9). Use math.comb() for cleaner code or calculate manually using factorials.