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Thread: Permutations/Combinations

  1. #1
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    Permutations/Combinations

    If upper-case and lower-case letters are considered as different litters, how many six-letter computer passwords are possibile with no repeated letters?

    I thought it was 52 pick 6, but its not, why wouldn't it be 52x51x50x49x48x47?
    Explanation would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by peekay View Post
    If upper-case and lower-case letters are considered as different litters, how many six-letter computer passwords are possibile with no repeated letters?
    I thought it was 52 pick 6, but its not, why wouldn't it be 52x51x50x49x48x47?
    You are correct the answer is $\displaystyle 52\cdot 51\cdot 50\cdot 49\cdot 48$$\displaystyle \cdot 47$.
    But that is not fifty-two choose (pick) six: $\displaystyle \binom{52}{6}=\frac{52!}{(6!)(46!)}$.
    The second is a combination, whereas the first is permutation.
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  3. #3
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    Isn't 52 pick 6 the same as 52x51x50x49x48x47?

    If you punch both into your calculator you get the same answer
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  4. #4
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    Quote Originally Posted by peekay View Post
    Isn't 52 pick 6 the same as 52x51x50x49x48x47?
    Absolutely not!
    Forget your calculator!
    Learn some basic mathematics.
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  5. #5
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    Learn some basic mathematics?...

    52 pick 6 = 52! / (52-6)!
    = 52! / 46!
    correct?
    52!/46!
    is the same as saying 52x51x50x49x48x47
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  6. #6
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    Quote Originally Posted by peekay View Post
    Learn some basic mathematics?...
    52 pick 6 = 52! / (52-6)!
    = 52! / 46!
    correct?
    52!/46!
    is the same as saying 52x51x50x49x48x47
    Absolutely wrong!
    Fifty-two choose (pick is a crude way to put it) six is:
    $\displaystyle \binom{52}{6}=\frac{52!}{(6!)(46!)}=\frac{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47 }{6\cdot 5\cdot 4\cdot 3\cdot 2 \cdot 1}$

    Why have you not learned the definitions involved in these questions?
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  7. #7
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    It's not choose, its pick.

    Pick refers to permutations while Choose refers to combinations. Pick is way different than choose.

    I have several examples in my textbook and its gives me the formula
    nPr = n! / (n-r)!

    Heres an example:

    In a card game, each player is dealt a face down "reserve" of 13 cards that can be turned up and used one by one during the game. How many different sequences of reserve cards could a player have?

    Solution
    Here, you are taking 13 cards from a deck of 52.

    52Pick13 = 52! / (52-13)!
    =52!/39!
    =52x51x50x...x41x40
    =3.9543x10^21


    Straight from the textbook
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  8. #8
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    Good grief! Those math-education folks have struck again.
    This distinction goes against 75 years of tradition.
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