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Math Help - How many eight-bit strings contain exactly three 0's

  1. #1
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    How many eight-bit strings contain exactly three 0's

    How many eight-bit strings contain exactly three 0's?

    Up till now, I've been getting problems right and I don't know why. I've just understood the method somehow. The method of writing:

    _ _ _ _ _ _ _ _

    And then writing how many possible number choices on the line. For this problem, I think any 3 could have a 0. So, I let the first three have two choices, and the other five with 1 choice.

    So, 2 x 2 x 2 x 1 x 1 x 1 x 1 x 1 = 8.

    That doesn't look right.

    Is there a proper method for probability?

    Thanks!
    Last edited by mr fantastic; December 3rd 2010 at 11:24 AM. Reason: Re-titled.
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  2. #2
    Senior Member BAdhi's Avatar
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    I assume that what you have mentioned as "bit" is binary numbers(am I correct?)

    lets take one situation where there is three zero's

    0,0,0,a,b,c,d,e

    a,b,c,d and e can either be 1 or 0. So all possible combinations for arranging those would be 2^5.

    just like the above situation there can be \frac{8!}{3!} situations

    so all possible combinations =  \frac{8!}{3!}.2^5

    from this you can get the probability of getting a string containing exactly three 0's
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  3. #3
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    How can a,b,c,d,e be 0 or 1?

    There can only be 3 zeros, which you put before a,b,c,d,e

    I thought I recetnyl had:

    _1_1_1_1_1_

    There are six positions where the three zeros could go.
    How I translate that into an equation...idk. Guessing: 2^6
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  4. #4
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    you are right
    hint - how many ways can you permute 11111000?
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  5. #5
    Senior Member BAdhi's Avatar
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    sorry, my mistake

    you are correct

    when there are three zeros there cannot be more zeros included.

    as "aman_cc" has mentioned the answer is the number of permutations in the group 11111000
    which is \frac{8!}{3!.5!}

    so the final answer would be \frac{8!}{3!.5!.2^8}

    am I correct now?
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  6. #6
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    Quote Originally Posted by BAdhi View Post
    as "aman_cc" has mentioned the answer is the number of permutations in the group 11111000 which is \frac{8!}{3!.5!}
    so the final answer would be \frac{8!}{3!.5!.2^8}
    The final answer is \dfrac{8!}{3!\cdot 5!}.
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  7. #7
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    Hello, Truthbetold!

    How many eight-bit strings contain exactly three 0's?

    We have eight spaces to fill with three 0's and five 1's:. _ _ _ _ _ _ _ _

    Select 3 of the 8 spaces: . \displaystyle _8C_3 \:=\:{8\choose3} \:=\:\frac{8!}{3!\,5!} \:=\:56 choices.

    Place 0's in those three spaces.
    . . Place 1's in the remaining spaces.


    Therefore, there are 56 eight-bit strings with exactly three 0's.

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  8. #8
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    Quote Originally Posted by BAdhi View Post
    sorry, my mistake

    you are correct

    when there are three zeros there cannot be more zeros included.

    as "aman_cc" has mentioned the answer is the number of permutations in the group 11111000
    which is \frac{8!}{3!.5!}

    so the final answer would be \frac{8!}{3!.5!.2^8}

    am I correct now?
    No, this is the number of 8 bit strings that have exactly 3 0s divided by the number of all 8 bit strings. It would be the probability that, if all 8 bit strings were equally likely to be chosen, of choosing a string that has exactly 3 0s.
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