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Math Help - Combinations (NcR)

  1. #1
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    Combinations (NcR)

    is this true for all k?

    k(n on top of r) = (kn on top of kr)
    if yes, explain why, if no, isit true for any k?

    NB: when i say 'on top of' i mean this kind of 'on top of' Combination - Wikipedia, the free encyclopedia
    the big bracket with the floating letters inside with no lines

    my first answer was k can be anything but 0, but then because 0! is 1 so my answer now is k can be any number, including neg numbers, what are your thoughts? thanks.
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  2. #2
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    Quote Originally Posted by LeiaBrown View Post
    is this true for all k?
    k(n on top of r) = (kn on top of kr)
    if yes, explain why, if no, isit true for any k?
    To be clear without wasting time.
    Are we assuming that k,~n,~\&~r are positive integers?

    And the question is k\binom{n}{r}=\binom{kn}{kr}~?

    BTW: [tex]k\binom{n}{r}=\binom{kn}{kr}~?[/tex] gives k\binom{n}{r}=\binom{kn}{kr}~?
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  3. #3
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    yes thats the question ^

    sorry i forgot "r=2"
    k, n and r are positive integers
    (and thanks for the [tex] thing)
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  4. #4
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    Quote Originally Posted by LeiaBrown View Post
    yes thats the question ^
    sorry i forgot "r=2"
    k, n and r are positive integers
    Well it does not work: 6\dbinom{10}{2}=240 and \dbinom{60}{12}=1399358844975
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    Quote Originally Posted by LeiaBrown View Post
    yes thats the question ^

    sorry i forgot "r=2"
    k, n and r are positive integers
    (and thanks for the [tex] thing)
    Is it true that \displaystyle 2 {3 \choose 1} = {6 \choose 2} ?
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    ok thanks, im still a bit confused but thanks anyway
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    Quote Originally Posted by LeiaBrown View Post
    ok thanks, im still a bit confused but thanks anyway
    What is there to be confused about? You've been given two examples (known as counter examples) of it not working .... Draw a conclusion.

    And Google Counter example.
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    Quote Originally Posted by LeiaBrown View Post
    ok thanks, im still a bit confused but thanks anyway
    You may not know how it works.
    \dbinom{10}{2}=\dfrac{10\cdot 9}{2\cdot 1}=45

    \dbinom{20}{4}=\dfrac{20\cdot 19\cdot 18\cdot 17}{4 \cdot 3\cdot 2\cdot 1}=5\cdot 19 \cdot 3\cdot 17.

    Does that help?
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  9. #9
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    OH I understand now, thanks a lot!!
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