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Permutations and Combinations

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Not really a problem, just my understanding.

    What is the difference between them? I know the formulae are different. They seem to be the same thing, that is, n objects taken r at a time. Any help in clarifying this would be appreciated.
     
  2. jcsd
  3. Dec 28, 2008 #2
    Permutations are all possible arrangements with the sequence of the objects considered. When the sequence of objects is not considered, then it results in a combination.
     
  4. Dec 28, 2008 #3
    Do you know about vectors and sets? Suppose you have a set of 3 objects, S = {1, 2, 3}. Permutations of 2 objects count the number of 2-object vectors you can construct from this set (without replacement), so you have the size of the set of vectors {(1, 2), (2, 1), (1, 3), (3, 1), (2,3), (3, 2)}, which is 6 distinct pairs. The amount of combinations of two objects formed from objects of this set is the size of the set of sets {{1, 2}, {1, 3}, {2, 3}} which is 3 distinct pairs. You count combinations when you are only concerned about which objects are in each bag or string of text for example, not the order in which they are in the bag. Permutations matter when you are concerned about the order in which the objects are placed in each bag or text string.
    Ie., if you are counting 3-letter words, combination counting would count the words DOT and TOD as the same word, while permutation counting notes that the letters are in a different order and counts them as different words.
     
  5. Dec 28, 2008 #4

    HallsofIvy

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    Suppose you have a club of 50 people and you want to choose a 3 people as "president", "vice president", and "secretary". You can choose any of the 50 people as president, then any of the 49 remaining people as vice president, then any of the 48 remaining people as secretary. There are, then 50*49*48 ways of doing that. One way to write that is
    [tex]50*49*48= \frac{50*49*48*\cdot\cdot\cdot*3*2*1}{47*46*\cdot\cdot\cdot*3*2*1}= \frac{50!}{47!}= \frac{50!}{(50-3)!}[/tex]. This is a "permutation" problem because
    "Bill for president, Mary for vice president, Charles for secretary" as different from "Mary for president, Charles for vice president, Bill for secretary".

    If instead you want a three person "operating committee" then it doesn't matter which is chosen first, which second, or which third. Now we do not want to treat "Bill for president, Mary for vice president, Charles for secretary" as different from "Mary for president, Charles for vice president, Bill for secretary". Now, since order in not important, it is a "combinations" problem. Since there are 3! ways of writing any three things, to discount interchanging the same three people, we need to divide by 3!:
    [tex]\frac{50!}{47! 3!}= _{50}C_{3}[/tex].
     
    Last edited: Dec 29, 2008
  6. Dec 28, 2008 #5
    Halls I think you meant to put [tex] \frac{50!}{47!}[/tex]
     
  7. Dec 28, 2008 #6
    You guys are great. I completely get it now!
     
  8. Dec 29, 2008 #7

    HallsofIvy

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    Thanks. I had written "\fra " instead of "\frac"! I haved editted it now.
     
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