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Factorial rules

  1. Nov 2, 2012 #1
    hey. I've had no education in factorials specifically, but my professor is expecting us to already know this stuff...

    In a problem where factorials are included, it is claimed (2n+2)! = (2n+2)*(2n+1)*(2n)!. Shouldn't it be (2n+2)!= (2n+2)*2n! ?

    In addition, is there any difference between 2n! and (2n)! ?
    Last edited: Nov 2, 2012
  2. jcsd
  3. Nov 2, 2012 #2


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    1. "(2n+2)!= (2n+2)*2n! ?" No, why?

    2. "In addition, is there any difference between 2n! and (2n)! ?"

    2*n! is twice the value of n!
    (2n!) is the factorial up to the number 2n

    For example, 2*3! =12, whereas (2*3)!=6!=720
  4. Nov 2, 2012 #3


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    What is a factorial? n! is defined as being all the positive integers up to and including n being multiplied together, so n! = n(n-1)(n-2)...3*2*1

    Ok, using this idea, what would (2n+2)! be? Well, first we multiply by (2n+2), then we reduce the value by 1 and multiply by that, so we multiply by ((2n+2)-1) = (2n+1).

    2n! = 2*(n!) = 2*(n(n-1)(n-2)...3*2*1)

    (2n)! = (2n)(2n-1)(2n-2)...*3*2*1
  5. Nov 2, 2012 #4
    1. Doesn't (2n+2)! = (2n+2)*(2(n-1)+2)*(2(n-2)+2)!

    2. Ah yes. How could I miss that? damn, i guess i'm exhausted.

    ah, OK. thanks 4 the help
  6. Nov 2, 2012 #5


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    I have a feeling you might be getting this mixed up with the techniques you learnt in Mathematical induction?

    Let's give n a value, say, n=5

    (2n+2)! = (2*5+2)! = 12! = 12*11*...*3*2

    Now, 2(n-1)+2 = 2*4+2 = 10.
    Notice how 10 is 2 less than 12, because we didn't take 1 away from the value 12, we took a value away from n, which is being multiplied by 2, so if we followed what you wrote we'd end up with 12! = 12*10*8*6*4*2

    Essentially, if we have a linear equation an+b for some constants a and b, then [itex]an+b-1 \neq a(n-1)+b[/itex] unless a = 1.
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