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N factorial - general question

  1. Mar 2, 2010 #1
    Hello everyone!

    1. The problem statement, all variables and given/known data
    n!/(n-r)! = n(n-1)(n-2)...(n-r+1)

    where r is the number of objects we want from n distinct objects (3 billiard balls out of 16)

    I don't understand what the last term in the expansion means, the (n-r+1). For example, suppose we have 5 distinct objects and we want to choose 2. This means that n = 5 and r = 2


    is this correct?

    Thank you for your help
  2. jcsd
  3. Mar 2, 2010 #2


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    Gold Member

    16!/(16-3)! = 3360 which is P(n,r)

    If you want C(n,r)=P(n,r)/r!
    Last edited: Mar 2, 2010
  4. Mar 2, 2010 #3
    The notation n(n-1)(n-2)...(n-r+1) actually means to start at n and multiply all the numbers below it until we reach n-r+1.

    Therefore with your example where n=5 and r=2, we have that
    n-r+1 = 5-2+1 = 4
    and we multiply
    5 * 4 = 20
    where we don't include any factor below 4.

    Similarly, if we had n=5 and r=3, then we'd find
    n-r+1 = 5-3+1 = 3
    so when we evaluated n(n-1)(n-2)...(n-r+1), we'd have
    5 * 4 * 3 = 60
    (once again, note that we don't include any factor below n-r+1 = 3)

    Hope that clarifies things.


    Additionally, like cronxeh mentioned, that expression counts permutations. In your example of the billiard balls, we'd want to count combinations, unless you're also giving some order to the choice.
  5. Mar 2, 2010 #4
    Awesome, thanks guys.
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