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Math Help - [SOLVED] counting and bit strings, just want to check my answer

  1. #1
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    [SOLVED] counting and bit strings, just want to check my answer

    How many bit strings of length six either begin with two 0s or end with three 0s?

    This tells me that there are only two option available to begin with, 0 or 1.

    For it to start with 2 0's, it would be 2^4=16
    For it to end with 3 0's, it would be 2^3=8
    Subtract 2^1 for the two instances that would have two 0's in the front and three 0's in the back.


    So: 16+8-2=22

    Is that correct?
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  2. #2
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    Yes it is correct.
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  3. #3
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    Plato,
    Would you mind helping me out on this one?
    Suppose a department contains 5 men and 7 women. How many ways are there to form a committee with four members if there needs to be at least one man and at least one woman?
    Would I set it up similar to this:
    C(5,1)*(7,1)= (5!/1!4!)*(7!/1!6!)

    I saw a similar problem:
    C(9,3)*(11,4)= (9!/3!6!)*(11!/4!7!)=84*330=27720

    What I dont get about it is, how did they get the 84*330?? That's the step I need to figure out my problem.
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  4. #4
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    Quote Originally Posted by ninobrn99 View Post
    Plato,
    Suppose a department contains 5 men and 7 women. How many ways are there to form a committee with four members if there needs to be at least one man and at least one woman?
    \sum\limits_{k = 1}^3 {\binom{5}{k}\binom{7}{4-k}}
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  5. #5
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    oooh...summations...umm, yeah... gonna have to look into that one thanks
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  6. #6
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    Well look at this this.
    \binom{5}{4} is the number of ways to have all men.
    \binom{7}{4} is the number of ways to have all women.
    \binom{12}{4} is the number of way to select period.
    So what is the number of ways to have at least one of each?
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  7. #7
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    Quote Originally Posted by Plato View Post
    Well look at this this.
    \binom{5}{4} is the number of ways to have all men.
    \binom{7}{4} is the number of ways to have all women.
    \binom{12}{4} is the number of way to select period.
    So what is the number of ways to have at least one of each?
    I gotta be honest with ya. I just can't remember how to do summations. I gotta reread that section.
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  8. #8
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    let me see if I got it:
    for all men, you have 120 ways
    for all women, you have 840
    for 12, its 11880

    i subtract the number 960 from that then divide by 24 to get 455?
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  9. #9
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    \binom{5}{4}=5 is the number of ways to have all men.
    \binom{7}{4}=35 is the number of ways to have all women.
    \binom{12}{4}=495 is the number of way to select period.
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  10. #10
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    Please check this

    Is this solution correct.
    Attached Thumbnails Attached Thumbnails [SOLVED] counting and bit strings, just want to check my answer-picture-1.png  
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  11. #11
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    Quote Originally Posted by alexriverajr View Post
    Is this solution correct.
    Thank you for trying. However, please try to be correct.
    \binom{7}{4}=\frac{7!}{4!(3!)}=\frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1} =35.
    Giving an incorrect answer is not helpful!
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  12. #12
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    Thanks

    Thanks for catching my mistake.
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  13. #13
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    Change

    Is this correct, i'm just looking for confirmation, not posting the answer.
    Attached Thumbnails Attached Thumbnails [SOLVED] counting and bit strings, just want to check my answer-picture-1.png  
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