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

The **Answer to the Question**

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**Here's the Solution to this Question**

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

$32-8=24$