Final answer:
The number of bit strings containing exactly five 0s and fourteen 1s with each 0 followed by at least two 1s is found using combinations, yielding 2002 unique bit strings.
Step-by-step explanation:
To solve how many bit strings contain exactly five 0s and fourteen 1s with the constraint that every 0 must be immediately followed by at least two 1s, we can apply combinatorial methods.
First, since each 0 must be followed by two 1s, we can treat each '0101' sequence as a single entity.
This leaves us with five '0101' entities and nine additional 1s to distribute, so we have a total of fourteen items to arrange (five '0101' entities and nine 1s).
Now, we need to find the number of ways to arrange these fourteen items.
This can be calculated using the formula for combinations:
C(n, k) = n! / (k! (n-k)!), where n is the total number of items and k is the number of items of one type.
In our case, n is 14 and k is 5.
The calculation gives us C(14, 5) ways to arrange the bit strings.
Using the combination formula:
C(14, 5) = 14! / (5! * (14-5)!)
= 2002
Therefore, there are 2002 different bit strings that meet the conditions provided.