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Permutations v. Combinations

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  1. Dec 8, 2014 #1
    We're currently studying counting and the different equations which are the foundation of probability (permutations and combinations). I understand that permutations are used when order is taken into account and combinations is for when order doesn't matter.

    However, the two equations appear very similar. I don't see how permutations can accommodate for order when combination cannot.
     
  2. jcsd
  3. Dec 9, 2014 #2

    Stephen Tashi

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    Which two equations are you talking about?
     
  4. Dec 9, 2014 #3

    DrClaude

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    I guess you mean
    \begin{align*}
    {}_nC_r &= \frac{n!}{r! (n-r)!} \\
    {}_nP_r &= \frac{n!}{(n-r)!}
    \end{align*}
    Notice that ##{}_nP_r = {}_nC_r \times r!##. Basically, the number of permutations of ##r## objects picked from ##n## is the number of combinations of ##r## objects picked from ##n## times the number of possible arrangements of ##r## objects, namely ##r!##.
     
  5. Dec 9, 2014 #4
    Yes, DrClaude those are the equations I was referring to. Do you have a quick concrete example? Maybe using a deck of a cards where I can actually see it for myself.
     
  6. Dec 9, 2014 #5

    DrClaude

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    It will be easier to use the letters of the alphabet. You are asked to pick three (without replacement). How many different outcomes can you get? You have 26 possibilities for the first letter, then 25 for the second, then 24 for the third, so ##26 \times 25 \times 24 = 15600##. This is the number of possible permutations, as you have taken the possibility of ABC as distinct from BAC, etc. Indeed,
    $$
    {}_{26}P_{3} = \frac{26!}{(26-3)!} = \frac{26!}{(23)!} = 26 \times 25 \times 24 = 15600
    $$
    How many possible permutations of ABC are there? You have 3 possibilities for the first letter, then 2 for the second, and then 1 for the last, sp ##3 \times 2 \times 1 = 3!##. So of all the permutations, ##3! = 6## correspond to the same three letters. To find the number of combinations, you have to divide by ##3!##. You get
    $$
    {}_{26}C_{3} = \frac{26!}{3! (26-3)!} = \frac{26!}{3!(23)!} = \frac{26 \times 25 \times 24}{6} = 2600
    $$
     
  7. Dec 9, 2014 #6

    PeroK

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    Try dealing 2 cards from a pack. If the order "matters", then the number of "permutations" is 52x51. Example: 6 of Hearts and King of Clubs is different (permutation) from Kings of Clubs and 6 of Hearts.

    But, if the order does not matter (which is usually the case with card games: your hand is the same no matter what order you got dealt the cards), then there are only 52x51/2 combinations.

    Example: 6 of Hearts and King of Clubs is the same hand (combination) as King of Clubs and 6 of Hearts.
     
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