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Homework Help: Permutations : hard question !

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    In how many ways can the three letters from the word " GREEN " be arranged in a row if atleast one of the letters is "E"

    2. Relevant equations

    Permutations Formula

    3. The attempt at a solution

    The total arrangements without restriction: 5P3/2! = [itex]\frac{5!}{2! * 2!}[/itex]

    The number of arrangements in which there is no "E" = 3!

    Ans : 5P3/2! - 3! = 24 (wrong)

    Here is another approach :

    The arrangements with just one "E" = [itex]\frac{2!*4!}{2!}[/itex]
    The arrangements with two "E" = [itex]\frac{3*2}{1}[/itex]

    I think I making a mistake due to the repetition of "E" ... Can any one of you tell me a better way which avoids the problem I am being having .
  2. jcsd
  3. Feb 6, 2013 #2


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    Science Advisor

    If "at least one letter must be E", that means that the other two letters must be chosen from "GREN". That is, the first letter must be one of those 4 letters, the second one of the three remaining letters. How many is that?

    But then we can put the "E" that we took out into any of three places: before the two, between them, or after the two letters so we need 3 times that previous number.
  4. Feb 6, 2013 #3
    4P2 = 4*3

    According to your method the answer should be 12 * 3 = 36 (The correct answer in the solutions is "27")

    Clearly your method (as did mine) repeats some of the permutations :

    You didn't take onto account that the two "E" are not distinct. Thus in those permutations where we chose two "E" were repeated. see :
    GEE , EEG, EGE
    REE, ERE , EER
  5. Feb 6, 2013 #4


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    Gold Member

    Alternatively, split the problem into two: Count those combinations with only 1 E and then count separately those with two E's.

    For one E combination: (1C1)*(3C2)*3! = 18
    For two E combination: (2C2)*(3C1)*(3!/2!) = 9. Then add.
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