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Permutation/Combination

I am not sure how to approach this question. I need help with all three letters, here is what I have so far: 1 character = 8 bit..

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    for a, there are two numbers 0 and 1 that can be in the sequence.. so 2^8 because there are 8 characters and 2 numbers 0 and 1 that could be possible? – 2017-01-27
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    Yes, that is right for a. For b, how many bits can you specify freely? How many are forced by other bits? For c, again how many bits can you specify freely? How many get forced by the requirement to have an even number of $1$s? – 2017-01-27
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    @RossMillikan is A right because of my guess at 2^8 or because my explanation is actually correct? – 2017-01-27
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    Your explanation is basically correct. I would not use the word characters as that is what you are storing. I would use bits, of which you have $8$. Each one gives you two choices independent of the others, so you multiply them to get $2^8$ – 2017-01-27
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    @RossMillikan no clue as to how to do b and c – 2017-01-27

1 Answers 1

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a) is relatively straight forward. For the first bit, there are 2 possibilities: 1 and 0. For the first two, the possibilities are 11, 10, 01, and 00. 4 in total. You will find that as you continue this trend, the number of combinations equals $2^n$ where n is the length of the character.

b) is similar. A palindrome is the same forwards as backwards. Therefore the last 4 numbers equal the first 4 in reverse order. Now there are only 4 digits to work with. Use the same system in a) as here, just with 4 instead of 8.

c) is a bit tricky. For that one, you simply reduce the power by 1. So with an 8 bit binary string with only even amounts of 1s, you would result with $2^{n-1}$ possibilities. In this case, $2^7$.

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    The point for c is that you choose the first seven bits freely, then the last bit has to be right to make the number of $1$s even and it can always do so. – 2017-01-27
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    @RossMillikan so is b 2^16 or 2^4, I can't figure that one out :( – 2017-01-27
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    @keving It would be 2^4. You get to choose the first 4 bits, and the last 4 have to match the first 4. – 2017-01-27
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    how can I get better at this? this seems so difficult but when I get the answer, seems simple .-. – 2017-01-27
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    @keving make yourself comfortable with multiplication principle. Most elementary counting questions can be broken down into either multiplication principle or addition principle. Be sure that when trying to use multiplication principle that you in fact count every outcome and you count every outcome exactly once (*or for more advanced questions, exactly the same number of times and divide by that number to account for the overcounting*). – 2017-01-27
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    @keving Combinatorics is a great subject because it requires a completely different mindset than most math you'll have met before. Rather than just grinding out some calculations, you almost have to find *exactly the right way* of looking at the problem (sometimes there are a couple good viewpoints, sometimes not). You just need practice getting familiar with the techniques you have at your disposal, and you build up a sense of which approach is the best for a problem. – 2017-01-27