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Math Help - combinations

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    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



    Attachment 3530
    (a) is just evaluating each:

    C(7,5) = \frac {7!}{5! ( 7 - 5)!} = \frac {7!}{5!2!} = \frac {7 \cdot 6}{2} = 21

    C(4,4) + C(5,4) + C(6,4) = \frac {4!}{4! (4 - 4)!} + \frac {5!}{4! (5 - 4)!} + \frac {6!}{4! (6 - 4)!}

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

    = 1 + 5 + \frac {6 \cdot 5}{2}

    = 1 + 5 + 15

    = 21

    (b)
    The second method is doing what i did for question two in your previous post, but at the end change C(5,1) to C(5,4) and C(6,2) to C(6,4), why can we do that?

    C(n,r) = C(n, n - r)

    so, for example, C(6,2) = C(6, 6 - 2) = C(6,4)
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  2. #17
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    Thankyou!!!!!!!!!!

    Thankyou To All Of Those People In Math Forum Who Heped Me In My Time Of Need.

    You Have My Eternal Gratitude, I Just Wanna Know How You Guys Do So Good In Math..........is There A Secret??!!!!!!because Even I Do My Homework And I'm Having Trouble Studying For My Grade 12 Geometry Exam!!!!!!
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raiden_11 View Post
    Thankyou To All Of Those People In Math Forum Who Heped Me In My Time Of Need.

    You Have My Eternal Gratitude, I Just Wanna Know How You Guys Do So Good In Math..........is There A Secret??!!!!!!because Even I Do My Homework And I'm Having Trouble Studying For My Grade 12 Geometry Exam!!!!!!
    Aha! i knew there was a more efficient way to do this. we can use Pascal's formula. i think it's better for you to do (b) that way, as well as question 2 that i did earlier maybe (if you want to).

    By Pascal's Formula, C(n,r) = C(n - 1,r - 1) + C(n - 1, r)

    So, we have:

    C(7,5) = C(6,4) + C(6,5) .........now expand the last combination using the formula

    \Rightarrow C(7,5) = C(6,4) + C(5,4) + C(5,5)

    Now, C(5,5) = C(4,4) = 1

    \Rightarrow C(7,5) = C(6,4) + C(5,4) + C(4,4)

    as desired

    QED


    Now, wasn't that a lot nicer?!
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