How many ten bit strings contain five consecutive zeros or five consecutive ones?

Answer 1

Given:

• Length (L) = 10
• Characters: 0, 1, therefore Count (n) = 2
• 5 consecutive zeros or 5consecutiveones

I'll try to map out some logic for you. I'm sorry this took so many edits for me to get the correct answer. These problems can some times be a little tricky, and I just kept tripping up on the logic.:

First, calculate the maximum amount of possibilities, without any constants.

This step is optional, however it gives you a check on your final answer.

Length = 10

n = 2

Therefore, the maximum amount of possibilities without any constants is 2^10 = 1084. If you get a number higher than 1084, which in one of my edits I did, than your answer is obviously incorrect.

Case 1: Fiveconsecutivezeros

[0][0][0][0][0][?][?][?][?][?] ==> 2^5 possibilities (2 x 2 x 2 x 2 x 2)

[1][0][0][0][0][0][?][?][?][?] ==> 2^4 possibilities

[?][1][0][0][0][0][0][?][?][?] ==> 2^4 possibilities

[?][?][1][0][0][0][0][0][?][?] ==> 2^4 possibilities

[?][?][?][1][0][0][0][0][0][?] ==> 2^4 possibilities

[?][?][?][?][1][0][0][0][0][0] ==> 2^4 possibilities (1 possibility duplicated from case 2)

Case 2:

[1][1][1][1][1][?][?][?][?][?] ==> 2^5 possibilities (2 x 2 x 2 x 2 x 2)

[0][1][1][1][1][1][?][?][?][?] ==> 2^4 possibilities

[?][0][1][1][1][1][1][?][?][?] ==> 2^4 possibilities

[?][?][0][1][1][1][1][1][?][?] ==> 2^4 possibilities

[?][?][?][0][1][1][1][1][1][?] ==> 2^4 possibilities

[?][?][?][?][0][1][1][1][1][1] ==> 2^4 possibilities (1 possibility duplicated from case 1)

Special Cases (don't count these, they are counted twice in previous cases)

[0][0][0][0][0][1][1][1][1][1] ==> 1 possibility (counted in case 1)

[1][1][1][1][1][0][0][0][0][0] ==> 1 possibility (counted in case 2)

We need to subtract 2 from the final answer to account for this.

As you can see, the number of possibilities for each case are:

2^5 + 5(2^4) = 112

There are two cases, you can multiply that by 2 to get your answer.

2(2^5 + 4(2^4)) = 2(112) = 224 possibilities

Subtract the special cases:

224 possibilities - 2 possibilities = 222 possibilities

The answer to the question is:

There are 222 unique ten bit strings that contain five consecutive zeros or five consecutive ones.