Are you are talking about 12-bit binary strings (

**0**and

**1**) in every case?

If so then, then 4096 is correct.

a) 220 is correct but your formula is wrong.

b) is very wrong.

At least three 1's is the same as at most three 0's.

How many ways are there to have zero 0's, one 0,, two 0's, and three 0's? Add them up.

You could also find the number of strings that have three 1's, four 1's, ... , twelve 1's.

c) The only way to have equal numbers of 1's and 0's is to have six of each.