First calculated How many bit strings of length eight contain six consecutive 0's ?
A) 0 0 0 0 0 0 __ __ -----> 4
B) __ 0 0 0 0 0 0 __ -----> 4
C) __ __ 0 0 0 0 0 0 -----> 4
| A ∩ B | = 0 0 0 0 0 0 __ __ ∩ __ 0 0 0 0 0 0 __ = 0 0 0 0 0 0 0 __ -----> 2
| A ∩ C | = 0 0 0 0 0 0 __ __ ∩ __ __ 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 -----> 1
| B ∩ C | = __ 0 0 0 0 0 0 __ ∩ __ __ 0 0 0 0 0 0 = __ 0 0 0 0 0 0 0 -----> 2
| A ∩ B ∩ C | = 0 0 0 0 0 0 __ __ ∩ __ 0 0 0 0 0 0 __ ∩ __ __ 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 -----> 1
| A U B U C | = | A | + | B | + | C | - | A ∩ B | - | A ∩ C | - | B ∩ C | + | A ∩ B ∩ C|
= 4 + 4 + 4 - 2 - 1 - 2 + 1 = 12 - 5 + 1 = 8
total bit strings possible with length 8 = 28 = 256
bit strings of length eight do not contain six consecutive 0's = 256-8 = 248