Final answer:
The total number of bit strings of length 6 where the last bit is 0 and two consecutive 0s are not allowed is 9. This is determined by considering the conditions for the first five bits and adding the fixed last bit to each valid arrangement.
Step-by-step explanation:
We need to determine how many bit strings of length 6 can be formed given the conditions that the last bit is 0 and two consecutive 0s are not allowed. Let's break down the problem step by step.
Since the last bit must be 0, we effectively only need to consider the first 5 bits. The first bit can be either 1 or 0. However, if the first bit is 0, the second bit must be 1 to prevent two consecutive 0s.
Thus, there are the following possibilities for the first two bits:
For the third, fourth, and fifth bits, following either starting pair, a 0 must always be followed by a 1, while a 1 can be followed by either 0 or 1. This property can be used to build all valid strings.
Therefore, the valid combinations for the 5 bits (excluding the fixed last bit) are as follows:
- 10101
- 10110
- 10111
- 11010
- 11011
- 11100
- 11101
- 11110
- 11111
Adding a '0' at the end of each of these strings to comply with the last bit restriction, we obtain nine valid bit strings of length 6.
So, the total number of bit strings of length 6 that satisfy the given conditions is 9.