Question

For example, two-digit number have 4 combinations: 11, 12, 21, 22. Three-digit number have 8 combinations: 111, 112,…222.

• How to get number of combinations for number that have 4, 5, … 10 or more digits?

Thanks

P.S. This refers to the Delphi 🙂

Answer 1

The answer is 2^N, where N is the number of digits.

This is a purely mathematical problem, and concerns very basic combinatorics. It is easy to see why 2^N is the right answer. Indeed, there are two ways to choose the first digit. For each such choice, there are two ways to chose the second digit. Hence, there are 2×2 ways to chose a two-digit number. For each such number, there are two ways to add a third digit, making 2×2×2 ways to construct a three-digit number. Hence, there are

2 × 2 × ... × 2 = 2^N

ways to construct a N-digit number.

In Delphi, you compute 2^N by Power(2, N) (uses Math). [A less naïve way, which works for N < 31, is 1 shl N.]