How many bit strings of length 10 contain at least three 1s and at least three 0s?

Solution:

From the question it is given that strings of length 10 contain at least three 1s and at least three 0s.

The total length is = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024.

String with no ones = 10!/(0!(10 - 0!)) = 1

String with no zeros = 10!/(0!(10 - 0!)) = 1

String with one one = 10!/1!(10 - 1!)=10

Strings with one zero = 10!/1!(10 - 1!)=10

Strings with two ones = 10!/2!(10 - 2!)=45

Strings with two zeros= 10!/2!(10 - 2!)=45

1024 - 1 - 1 - 10 - 10 - 45 - 45 = 912.

Therefore, 912 strings of length 10 contain at least three 1s and at least three 0s.

How many bit strings of length 10 contain at least three 1s and at least three 0s?

Summary:

912 strings of length 10 contain at least three 1s and at least three 0s.