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Math Help - combinations

  1. #1
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    combinations


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  2. #2
    Bar0n janvdl's Avatar
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    Why does that look like an online test?
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  3. #3
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    Show that C(8,4)=C(7,3)+C(7,4)

    Let's manipulate the left side to become the right side
    \frac{8!}{4!\cdot (8-4)!}=\frac{8!}{4!\cdot 4!}=\frac{8\cdot 7!}{4!\cdot 4\cdot 3!}=\frac{2\cdot 7!}{4!\cdot 3!}=\frac{7!}{4!\cdot 3!}+\frac{7!}{4!\cdot 3!}
    \frac{7!}{3!\cdot (7-3)!}+\frac{7!}{4!\cdot (7-4)!}=\boxed{C(7,3)+C(7,4)} which was to be shown.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    Why does that look like an online test?
    It probably is. Or online homework anyway.

    -Dan
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    It probably is. Or online homework anyway.

    -Dan
    Usually you dont get points for homework...
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by janvdl View Post
    Usually you dont get points for homework...
    you do if the professor decides to make your homework grade a part of your final grade


    4) \sum_{i = 1}^{5} i \left[ \sum_{j = 1}^{3} (2j + 1) \right] = (1 + 2 + 3 + 4 + 5) \left[ (2(1) + 1) + (2(2) + 1) + (2(3) + 1) \right]

    = 15 ( 3 + 5 + 7)

    = 15 \cdot 15

    = 225
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  7. #7
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    What about the other ones?

    What about #2 and #3?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    What about #2 and #3?
    #2 is very similar to #1 which rualin so graciously did for you. try it on your own and tell us what you come up with.

    for problem #3, i'd do something like this. chances are there's a more efficient way, but this is how i saw the solution:


    \sum_{j = 3}^{6} (2j - 2) = \sum_{j = 1}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 2)

    = \sum_{j = 1}^{4} (2j - 2) + \sum_{j = 5}^{6} (2j - 2) - \sum_{j = 1}^{2} (2j - 1) .......evaluating the last two summations we obtain

    = \sum_{j = 1}^{4} (2j - 2) + 16 ........now let's do some manipulation on the first summation

    = \sum_{j = 1}^{4}(2j + 2 - 4) + 16 .......i didn't change anything, +2 - 4 = -2

    = \sum_{j = 1}^{4} (2j + 2) + \sum_{j = 1}^{4} (-4) + 16 ......now evaluate the second summation

    = \sum_{j = 1}^{4} (2j + 2) - 16 + 16

    = \sum_{j = 1}^{4} (2j + 2) ........a bit more manipulation on this summation should do it

    = \sum_{j = 1}^{4} (2j + 4 - 2) .........again, I changed nothing, +4 - 2 = +2

    = \sum_{j = 1}^{4} \left[ 2(j + 2) - 2 \right] .........i factored out a 2 from the first two terms

    and now we have the desired summation

    QED


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  9. #9
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    Are u sure?

    Are u sure there aren't any more steps?
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  10. #10
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    I still don't know?

    I still don't know how to do #2.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    Are u sure there aren't any more steps?
    more steps for what?

    you should click the "quote" button when you're responding to something, so people know exactly what you're responding to
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  12. #12
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    What i mean is....

    Quote Originally Posted by Jhevon View Post
    more steps for what?

    you should click the "quote" button when you're responding to something, so people know exactly what you're responding to


    Are you sure there aren't anymore steps for number 3....
    also i still don't know how to do number 2..
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    I still don't know how to do #2.
    here's one way to do it:

    C(7,5) = \frac {7!}{5! ( 7 - 5)!}

    = \frac {7!}{5! 2!}

    = \frac {7 \cdot 6!}{5! 2!}

    = \frac {(2 + 5) \cdot 6!}{5! 2!}

    = \frac {2 \cdot 6! + 5 \cdot 6!}{5! 2!}

    = \frac {2 \cdot 6!}{5! 2!} + \frac {5 \cdot 6!}{5! 2!}

    = 6 + \frac {6!}{4! 2!}

    = 1 + 5 + \frac {6!}{4! 2!}

    = C(4,4) + C(5,1) + C(6,2)

    QED
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    Are you sure there aren't anymore steps for number 3....
    also i still don't know how to do number 2..
    no, number 3 is complete. i started with one series and manipulated it and ended up with the other series. this proves the claim that they are equal. in fact, i think i probably did too many steps, i'm not very efficient with this sort of thing.
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  15. #15
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    This is my final question for math forum!!

    This is my final question for math forum which i hope you can answer!

    1. On page 345 of the textbook, the diagonal pattern of Pascalís Triangle illustrates that C(7, 5) = C(4, 4) + C(5, 4) + C(6, 4). Prove this relationship using the following methods.

    a)numerically by using factorials
    b)by reasoning, using the meaning of combinations



    combinations-pascals-20triangle.jpg
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