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Permutations problem

  1. Jan 28, 2007 #1
    Hi,

    I have to find the total number of permutations of four letters that can be selected form the
    word "ARRANGEMENT".

    Clearly we have 7 different letters so the amount of 4 letter permutations with no repeats is:

    7!/3!=840

    now for each two letter can form a four letter permutaion with another two different letters

    4!/(2!*2!)*6*5

    where the first part gives the number of permutations of a four letter word with 2 letters the same. Each of the remaining slots can take one of the other 6 letters and the other by one of the remaining 5 letters.

    Now given that there are 4 of these 180*4=720

    Finally must look at all the combinations of the double letters to form a four letter permutation:

    4!/(2!*2!)*3*4=72

    The first part is te number of permutations of a given two letters within a four letter sequence. This is then multiplied by the number of reminaing double letters it may forma permutation with 3. This total is then multiplied by 4 the total number of double letters as any oneof them could form the inital set of permutations.

    so I get a total of 72+720+840=1632.

    However correct answer is 1596 can someone please explain?
     
    Last edited: Jan 29, 2007
  2. jcsd
  3. Jan 30, 2007 #2

    ssd

    User Avatar

    A,R,N,E occur twice and G,M,T occur once.

    1/ No. of arrangement when all letters different: =7x6x5x4=840

    2/ No. of arrangement when 2 are of one kind and others different:=
    (no. of ways to select the letter to be repeated) x (no. of ways to select the letters not to be repeated) x (no. of ways to arrange them)
    4C1 x 6C2 x 4!/2! = 4 x (6x5/2) x 4!/2! =720
    [6C2= 6 combination 2]

    3/ No. of arrangement when two letters used (each repeated twice):=
    4C2 x 4!/(2!x2!) = 36

    840+720+36=1596
     
    Last edited: Jan 30, 2007
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