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Permutation problem

  1. Aug 18, 2007 #1
    In how many ways the letters in the word PARALLEL can be arranged so that there will always be two "L"s together as in the original word?
    I have asked this question in yahoo answers. None of them have come out with the right answer. Amazing! It's such an easy problem.:eek:
  2. jcsd
  3. Aug 18, 2007 #2


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    if you partition the letters into cells of two places, you get that there are four option to arrange the LL, now after you arrange the LL, you have 6 more letters to arrange in how many ways?
    after that use the multiplication theorem.

    wait, a second are you asking for help, or are you just giving us an example of yahoo stupidity?
  4. Aug 18, 2007 #3
    my answer is 7!/2-6!/2=2160
    I will be very glad if you carefully verify the answer.
  5. Aug 18, 2007 #4


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    if you first choose the LL where to place then you have 4 ways to do it, and two ways to to arrange the LL, after that you have six letters from which you need to sit them in the other 6 places, i think that it doesnt matter that one cell is taken you can still think of the problem of sitting the 6 letters as sitting 6 letters in a row, how many ways are there for this problem?
    combining them all by multiplication you get your answer, or so i think. (-:
  6. Aug 18, 2007 #5


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    Here's how I would do it: treat two of the L's as a single letter since they must be together. You now have 7 "letters", two of which, the As, are the same. The number of ways of writing 7 letter with two the same is 7!/2!= 5040.

    You don't give any reason why you subtract off 6!/2 so I can't comment on that. (My solution includes the possiblity that all 3 Ls will be together. I see nothing in the original question that excludes that.)
  7. Aug 18, 2007 #6


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    don't we have order in mind, i.e there are two ways of arranging the two Ls, and without the two Ls there're in overall 6 letters to arrange, if you also mind how As are arranged then you have as I said 4*2*6! ways to arrange, cause we have four ways to arrange the LL and for every arrangement of them we have 2 options to arrange every L in the pair, but if we don't care how we do it then we don't need two, and so we get 6!*4.
  8. Aug 18, 2007 #7


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    Yoyu're double-counting when there are three Ls together, though, right? That is, you count (LL)L as different from L(LL), when they're actually the same.
  9. Aug 19, 2007 #8
    PARALLEL has 8 letters with one letter appearing 3 times and 1 letter appearing 2 times.

    The letters can be arranged in 8!/(3!*2!) = 3360 ways.

    If you want 2 Ls together, you can treat it is 1 single letter.
    So it seems that the letters can be arranged in 7!/(2!) = 2520 ways.

    But in this case you are treating the letter LL and L to be different. Yes they are different when they are in separate place (i.e when LL and L are not together). But they are not always different.
    Consider the following cases
    (1) P A R (LL) A E L and P A R L A E LL (two letter (LL) and L has just exchanged their position). These two permutations are not identical. so LL and L is different letters here.

    But "PAR(LL)LAE" & "PARL(LL)AE" is identical though I have exchanged their position. So the letter (LL) and L are not same here. That infers the answer shouldn't be 7!/(2!*2!) = 1260 in which LL and L is considered identical in character.

    So I have to regard LL and L as identical letter when they are together and different when isolated. In how many permutations are they together? In 6!/2! permutation 3 Ls are together.
    If we could differ the third L from first two (LL) then there should be 7!/2! valid permutation (having 2 Ls together). But in these permutationS there are 2*6!/2! where 3 Ls are together and half of them is exactly same as the other half. So this half should be omitted to get the desired result.
    So the number of valid permutation is 7!/2!-6!/2! = 2160
    Last edited: Aug 19, 2007
  10. Aug 19, 2007 #9
    I checked out some of these smaller cases, and the formula ever works in the event of ALLL, where there are no cases excluded. Here we have 4!/3! = 4, and actually 3!-2! = 4, the correct answer. I rewrite the problem as AuL, where u=LL, and further rewrite it as As, where s=LLL.

    In AALLL, there are 5!/(2!3!) = 10 and the only excluded case is LALAL, so there are 9 cases involved. Using the formula: 4!/(2!)-3!/(2!) = 9, the correct answer. Now if we change that to ABLLL, lacking a double term in A, we have now 20 cases with 2 excluded, so correct anwer is 18. Changing this to ABuL, and ABs, and using formula, we get correct answer: 4!-3! = 24-6 = 18.

    Take the case of BAALLL, equal 60 cases. Here we can put the two A's in the middle and have BLALAL, and move the B into anyone of the five other places giving 6 excluded cases. But in the form ALBLAL, we get only three new cases, ALBLAL, LBLAAL, LBLALA and similarly if we exchange the B for the other internal A, ALALBL, LAALBL, LALBLA. So 48 is the right answer. 5!/2-4!/2 = (120-24)/2 = 48.
    Last edited: Aug 19, 2007
  11. Aug 21, 2007 #10
    Good examples!
  12. Aug 21, 2007 #11
    Thanks for the complement. I am glad you thought it was helpful.
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