How many eight-bit binary strings contain at least two 1's?

The number of 8-bit strings containing 2 1's =

Number of combinations of 2 from 8 = 8C2 = 8*7 = 28 strings

For 3 1's we have 8C3 = 56

For 4 1's we have 8C4 = 70

For 5 's it is 56, For 6 it is 28, For 7 it is 8 and for 8 it is 1

So our answer is

1 + 8 + 2 (28) + 2 (56) + 70 = 247 (answer)