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Math Help - Combinations - Problem with simplifying Factorials.

  1. #1
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    Combinations - Problem with simplifying Factorials.

    Hey folks, I'm having a problem with a combination question, hoping to get some ideas of where I'm going wrong from some pros!

    Sorry for the bad format, I usually use a program called MathType5 to type out my work, but I can't seem to get the code to translate over here, so I'm gonna do my best with microsoft word and show my work.

    Update, so sorry, I just can't get the fractions to translate over.

    Solve for "n" in n-1C11 = nC12

    I started by using the "n! / (n-r)!r!" formula on both sides, and tried to simplify from there by gathering like terms and then cross multiplying and dividing to get
    n!(n-2)!11! = (n-1)!((n-12)!12!)
    and then simplifying this into
    (n^2-12n)!11n! = (n^2-13n!)(12n-12)!
    And it is here that I am stuck, I don't know how to isolate n with factorials being attached to everything.


    Again sorry for the bad formatting, any help is appreciated and thanks for taking the time to look at my problem!


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  2. #2
    Moo
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    Hello,

    Uh-oh... a!b! \neq (ab)!

    I don't really know how you got to your answers though...

    Here is the beginning (click on the pictures to get the code)

    ^{n-1}C_{11}=^n C_{12}

    \frac{(n-1)!}{11!(n-12)!}=\frac{n!}{12!(n-12)!}

    You can see that (n-12)! cancels on each side :

    \frac{(n-1)!}{11!}=\frac{n!}{12!}


    Now note that n!=n\times(n-1)\times\dots\times 1=n\times (n-1)!

    Can you finish it ?
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  3. #3
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    That did it, I got n = 12.

    Thanks so much for the help!
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