Answer to Question #226032 in Discrete Mathematics for asha

Question

.Determine how many bit strings of length 5 can be formed ,where three consecutive 0s are not allowed .

Answer 1

The number of elements of the set of different binary strings of length 5 is $2^5=32.$

Determine the number of bit strings of length 5 that contain three consecutive 0s.

$$\begin{matrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 1\\ \\ 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 1\\ \\ 0 & 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0\\ \end{matrix}$$

How many bit strings of length 5 can be formed ,where three consecutive 0s are not allowed

$$32-8=24$$