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Question about permutations when n = k

  1. Aug 10, 2013 #1
    So the general formula for permutations as I understand it is n!/(n-k)!

    but what if n=k?

    so lets say you want to see how many ways you can seat 5 people in 5 chairs.

    then the answer would be 5!/(5-5)! which would be undefined...but logically it should be defined

    what did I misunderstand here?
     
  2. jcsd
  3. Aug 10, 2013 #2

    rubi

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    (5-5)! = 0! = 1 (by definition), so the term is well-defined and gives you 5! = 120.
     
  4. Aug 10, 2013 #3
    The number of ways to seat 5 people in 5 chairs is 5!.
    There are 5 choices of a seat for the first person, 4 for the second, 3 for the third...

    This follows the convention that ##0! = 1##.
     
  5. Aug 10, 2013 #4
    Oh, I did not realize that 0! = 1. I'm not really sure how 0! = 1 makes any sense, but yeah, logically I assumed the answer was just 5!, I was just confused by plugging numbers into the actual formula, how to divide by 0!
     
  6. Aug 10, 2013 #5

    rubi

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    It's just a definition. You define the factorial recursively by [itex]0! = 1[/itex] and [itex]n! = n (n-1)![/itex] for all natural numbers. With this definition, you can compute all the factorials, for example:
    [itex]3! = 3 \cdot 2! = 3 \cdot 2 \cdot 1! = 3 \cdot 2 \cdot 1 \cdot 0! = 3 \cdot 2 \cdot 1 \cdot 1 = 6[/itex]
     
  7. Aug 11, 2013 #6
    How many ways are there to arrange 0 objects? There is 1 way.

    ##3! = \frac{4!}{4} = \frac{24}{4} = 6##

    ##2! = \frac{3!}{3} = \frac{6}{3} = 2##

    ##1! = \frac{2!}{2} = \frac{2}{2} = 1##

    ##0! = \frac{1!}{1} = \frac{1}{1} = 1##
     
    Last edited: Aug 11, 2013
  8. Aug 11, 2013 #7
    The intuition behind there only being 1 way to arrange 0 objects makes sense....but what about the formula? Why is 4! equal to 5!/4? Also wouldnt that be an infinite loop, like 4!=5!/4=6!/5....etc?
     
  9. Aug 11, 2013 #8
    Nope. ##4!=\frac{5!}{5}=\frac{120}{5}=24## This is not a formula it's a generalisation.

    The actual formula is:

    ##n!=n\times(n-1)\times(n-2)\times(n-3)\times...(n-n)!##

    ##4!=4\times(4-1)\times(4-2)\times(4-3)\times(4-4)! = 24## and remember ##(4-4)! = 1##
     
    Last edited: Aug 11, 2013
  10. Aug 11, 2013 #9
    Right, it's just that using that actual formula I don't see how 0! = 1, but I see how it makes sense intuitively
     
  11. Aug 12, 2013 #10

    Remember we have established that there is ##1## way to arrange ##0## objects. So ##0!## really just means ##1## and if that is true then ##0!=1\times(1-1)!=1##
     
  12. Aug 12, 2013 #11
    Got it, so it's just a definition. Makes sense, thanks!
     
  13. Aug 12, 2013 #12

    jbriggs444

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    Yes, just a definition. But somewhat more general than you might expect.

    If you add up a list of numbers with no entries, the sum is taken as zero. Zero is the additive identity. This is sometimes called an "empty sum".

    If you multiply a list of numbers with no entries, the product is taken as one. One is the multiplicative identity. This can be called an "empty product".

    A list that starts at entry number 1 and ends at entry number 0 is empty. That is a convention that is adhered to in programming languages such as Ada.

    So the list of all the numbers from 1 on up to 0 is empty. If you multiply all the entries in that list together you get an empty product. An empty product is equal to 1.
     
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