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How many permutation of abcde are there in which confused!

  1. Oct 24, 2006 #1
    How many permutation of abcde are there in which .... confused!

    Hello everyone i'm lost on this problem, it says:

    How many permutations of abcde are there in which the first character is a, b, or c and the last character is c, d, or e?

    They say:
    The number of elements in a certain set can be found by computing hte number in some larger universe that are not in the set and subtracting this from the total.

    I'm not sure what "larger universe" i'm suppose to find..
    I nkow the inclusion/exculsion rule can be used to compute the number that are not in the set but the examples in the book don't seem to help me.


    If I let A = {a, b, c} and B = {c, d, e}
    then A U B = {a, b, c, d, e}
    A Insersect B = {c}
    N(A U B) = N(A) + N(B) - N(A intersect B); <-- thats the incuslion/exclusion rule...

    ANy help would be great
    thanks!
     
  2. jcsd
  3. Oct 25, 2006 #2

    StatusX

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    Homework Helper

    So what are the permutations that aren't of the form you want?
     
  4. Oct 25, 2006 #3

    HallsofIvy

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    Science Advisor

    I don't quite see how the "hint" applies either. I would do this directly:

    If the first letter is "a" then the last letter could be any of {c, d, e} and there are 3! ways of arranging the other 3 letters: 3(3!)= 18 ways.

    If the first letter is "b" then the last letter could be any of {c, d, e} and there are 3! ways of arranging the other 3 letters: again 18 ways.

    If the first letter is "c" then the last letter must be either "c" or "d" and there are 3! ways of arrranging the other 3 letters: 2(3!)= 12.

    A total of 18+ 18+ 12= 48 permutations.
     
  5. Oct 25, 2006 #4
    Thanks Ivy, I like your approach alot better than what they had in mind.
     
  6. Oct 25, 2006 #5
    whoops double post
     
  7. Oct 26, 2006 #6
    If you wanted to use the inclusion-exclusion principle you could do the following:

    Let,
    S = {a,b,c,d,e}
    X = 5-permutations of S
    A = 5-permutations of S with d or e as the first element
    B = 5-permutations of S with a or b as the last element.

    Note: I am going to use |X| to denote the cardinality of the set X, and A' to denote the complement of A.

    Note that this means A' and B' is what they are looking for. That is, [itex]A' \cap B'[/itex] = 5-permutations of S with a, b, or c as the first element, and c, d, or e as the last element (which is what we want to count the number of).

    Then using the inclusion exclusion principle we get [itex]|(A' \cap B')| = |X| - |A| - |B| + |(A \cap B)|[/itex]


    Now let us compute the cardinality of each set.

    |X| = 5!
    |A| = 2*4!
    |B| = 2*4!
    [itex]|(A \cap B)| = 2*2*3![/itex]

    So we get 5! - 2*4! - 2*4! + 2*2*3! = 48.
     
    Last edited: Oct 26, 2006
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