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)

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

486in is correct

step-by-step explanation: