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Math Help - Factorial Notation

  1. #1
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    Factorial Notation

    I keep having a hard time solving this factorial notation question. I know the steps but I'm having a brain fart of what the correct answer be to input into the next step.

    (n+2)! =110
    (n)!


    (n+2)x(n+1)x(n)(n-1)(n-2)....x1
    (n)x(n-1)x(n-2)....x1

    (n+2)(n+1)


    This is where I get confused.

    would the correct answer be n(squared)+3 = 110??

    Any help would be gratefully appreciated.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Factorial Notation

    What do you have to do? Find a n wherefore (n+2)!=110? or? ...
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  3. #3
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    Re: Factorial Notation

    Is the question
    (n + 2)! = 120 ???????
    The result should come "by inspection".

    Also, you should know a few factorials.
    0! = 1
    1! = 1
    2! = 2
    3! = 6
    4! = 24
    5! = 120
    etc
    So 5! = (n + 2)!
    5 = n + 2
    n = 3

    This is, of course, if the problem is 120 (not 110)!
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Factorial Notation

    Quote Originally Posted by TheChaz View Post
    Is the question
    (n + 2)! = 120 ???????
    The result should come "by inspection".

    Also, you should know a few factorials.
    0! = 1
    1! = 1
    2! = 2
    3! = 6
    4! = 24
    5! = 120
    etc
    So 5! = (n + 2)!
    5 = n + 2
    n = 3

    This is, of course, if the problem is 120 (not 110)!
    (x+2) facterial=110 solve for x - Wolfram|Alpha
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    Re: Factorial Notation

    I meant the first equation to say : (n+2)! / (n!) = 110
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    Re: Factorial Notation

    Quote Originally Posted by momofmaxncoop View Post
    I meant the first equation to say : (n+2)! / (n!) = 110
    \frac{(n+2)!}{n!}=\frac{(n+2)(n+1)...2\cdot 1}{n(n-1)...2\cdot 1}=(n+2)(n+1)


    (n+2)(n+1)=110
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    Re: Factorial Notation

    Quote Originally Posted by momofmaxncoop View Post
    I keep having a hard time solving this factorial notation question. I know the steps but I'm having a brain fart of what the correct answer be to input into the next step.

    (n+2)! =110
    (n)!
    Surely the question means \frac{(n+2)!}{n!}=110.
    That is
    \begin{align*}  \frac{(n+2)!}{n!}&= (n+2)(n+1)\\ (n+2)(n+1)&=110 \\  n^2+3n-108&=0 \end{align*}
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  8. #8
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    Re: Factorial Notation

    Where are you getting the -108??? shouldn't that be -110??
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    Re: Factorial Notation

    Quote Originally Posted by momofmaxncoop View Post
    Where are you getting the -108??? shouldn't that be -110??
    \begin{align*}(n+2)(n+1) &=110\\n^2+3n+2 &= 110\\ n^2+3n+2-110&= 0\\ n^2+3n-108&= 0 \end{align*}
    Last edited by Plato; August 17th 2011 at 01:22 PM.
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  10. #10
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    Re: Factorial Notation

    Can someone explain to me what you guys are doing in this problem? Are you just solving for n? If so then why isn't the answer just intuitively n=9?

    Nevermind, I think I get it now. After solving the polynomial I get n=9, 12

    But what's with the 12? It works with the polynomial but obviously not with the equation using factorials.

    Just curious
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  11. #11
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    Re: Factorial Notation

    Quote Originally Posted by downthesun01 View Post
    Can someone explain to me what you guys are doing in this problem? Are you just solving for n? If so then why isn't the answer just intuitively n=9?

    Nevermind, I think I get it now. After solving the polynomial I get n=9, 12

    But what's with the 12? It works with the polynomial but obviously not with the equation using factorials.

    Just curious
    The two solution are n=9, -12 which are both solutions to the quadratic but the second does not solve the original problem since n must be >=0.

    So when you get to the quadratic you are looking for a non-negative n which satisfies the quadratic. The solution to that problem is n=9.

    CB
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