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Permutation and Combination

  1. Aug 27, 2009 #1
    Hi guys, i have no idea how the Permutation should be used.

    An example, Find the number of arrangement of all nine letters of the word SELECTION in which

    a)the two letters E are next to each other
    Well i can solve this, i just make the EE as one unit so 8P8

    b)the two letters E are not next to each other
    I don't know how i should solve this, why couldn't i take 9P9 - 8P8 . It make sense to me, take away all those that the EE are together.

    Also when its 9P8 what does it means? Is it like there's 9 space and you gonna put 8 things?

    Because i was taught that 10C3 means 10 choose 3.
     
  2. jcsd
  3. Aug 28, 2009 #2
    Recall the formulas:

    [tex]_n C_r = \frac{n!}{(n-r)! \cdot r!}[/tex]

    [tex]_n P_r = \frac{n!}{(n-r)!}[/tex]

    The latter finds the number of ways of arranging in order r objects selected from n distinct objects.

    The issue with SELECTION is the two E's (which are not distinct).

    Fortunately there is a formula that finds the number of permutations of n objects of which n1 are of a 1st type, n2 are of a 2nd type, ...., n_k are of a kth type:

    [tex]\frac{n!}{n_1 ! \cdot n_2 ! \cdots n_k !}[/tex]

    You are correct to concatenate the E's in the first question. For the second, you need to find all of the ways of permuting all the letters and then remove from the total those arrangements where the E's are adjacent (which you determined in the first question).

    Without revealing anything, the answer to the second question is bigger than 100,00 and has only 4 different numerals in it.

    --Elucidus
     
  4. Aug 28, 2009 #3
    But mind telling me why is 9P9 - 8P8 wrong?
     
  5. Aug 29, 2009 #4
    9P9 is the number of ways of permuting 9 distinct objects. SELECTION does not consist of 9 distinct objects (there are two E's).

    There are actually 9!/2 ways of permuting all the letters in SELECTION.

    --Elucidus
     
  6. Aug 29, 2009 #5
    Actually i am in a better understanding of when to use C but not sure when to use P.

    Since SELECTION has 2 E's Why must i have 9!/2 ? I know 9! because the first place has 9 words to choose from and second place has 8 to choose from and so on. But why /2 ? Because there's 2 E's?
     
  7. Aug 29, 2009 #6
    Because there are 2! ways to arrange 2 E's. If you had a word with 3 E's, you'd divide by 3!. P gives you the number of ways where order matters, but C is for when order doesn't matter. Since there's no difference between EE and EE, you divide by 2!.
     
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