In how many ways a binary code consisting of 12 binary digits can be created, so that 10 or more digits are 1s?

This is a combination problem. The formula for combinations is C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

We want to find the number of ways to choose 10, 11, or 12 digits to be 1s out of 12 digits. So, we need to calculate C(12, 10) + C(12, 11) + C(12, 12).

1. C(12, 10) = 12! / [10!(12-10)!] = 66
2. C(12, 11) = 12! / [11!(12-11)!] = 12
3. C(12, 12) = 12! / [12!(12-12)!] = 1

So, there are 66 + 12 + 1 = 79 ways to create a binary code consisting of 12 binary digits so that 10 or more digits are 1s.