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# n-Bit Binary Strings with Fixed Digits

I need to show that the number of n-bit binary strings that contain exactly two occurrences of 10 is C(n+1, 5).

#### Solution Preview

Let's look at the number of ways to do this:

It's good to start with a simple example to gain insight into how the problem will work. For example, take some number of length n:

1010xxxxxx...... where the x's are the numbers which could be anything. We already have the two exaples of the '10', so no more can appear in the latter string of numbers.

So, if we have a string of length m, how many possible values are there which *don't* include 10? Well, once we place a '1' in the number, it can't be followed by any '0's, so they will be like:

11111.....
01111....
00111.....
00011....
.
.
00000...

Get the idea? If the string is of length m, then this means there are m+1 different ways to do it. Again, this is only when you can't have 10 ...

#### Solution Summary

It is shown how it may be proven that the number of n-bit binary strings that contain exactly two occurrences of 10 is C(n+1, 5).

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