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Thread: Permutation problems (circular & with repetition)

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    Question Permutation problems (circular & with repetition)

    Suppose we have 3 A's, 5 B's, 7 C's, 9 D's, and one E, F, G, and H.

    1. How many different arrangements of these letters are possible if I write them as a string?

    I believe this is 28! / (3! 5! 7! 9!). Seems pretty simple.

    2. If I write my "string" as a loop, either around in a circle or around the outside of a cylinder(so there is not just one beginning or end), how many different arrangements are possible?

    My attempt...

    For each permutation; there are 27 others that are considered to be identical?

    [28! / (3! 5! 7! 9!)] / 27.
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    Re: Permutation problems (circular & with repetition)

    Quote Originally Posted by Induction View Post
    Suppose we have 3 A's, 5 B's, 7 C's, 9 D's, and one E, F, G, and H.
    1. How many different arrangements of these letters are possible if I write them as a string?
    I believe this is 28! / (3! 5! 7! 9!). Seems pretty simple.

    2. If I write my "string" as a loop, either around in a circle or around the outside of a cylinder(so there is not just one beginning or end), how many different arrangements are possible?
    Your answer for #1 is spot on correct! Good for you.

    #2. Seat H at the circle. Now it is ordered!
    There are now $\dfrac{27!}{3!\cdot5!\cdot7!\cdot9!}$ ways to place the others.
    Thanks from Induction
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    Re: Permutation problems (circular & with repetition)

    Cheers.

    I only had one other example to work from and it went like this... in how many ways can six people be seated around a circular table?

    The solution says for each permutation (total of 6!); there are 5 others that are considered to be identical; so, 6! / 6 = 5!. Given your way; I suppose you could set one of the people and there are now 5! ways to place the others (I prefer this way; the other way is a bit confusing to me).

    Back to this question; I think I understand your approach. Set one of the "unique" elements (either E, F, G, H) in place. There are 27! permutations for the rest of the letters; but, we divide out by the permutations of the repeated letters (A, B, C, D). I'm understanding you with this, yes?

    I'd like to ask another question 'cause I like your approach. If every letter was repeated; what would happen? 28! would be divided out by the permutations of the repeated letters? Or is it still 27! divided out by the permutations of the repeated letters?

    Cheers for assistance.
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    Re: Permutation problems (circular & with repetition)

    Quote Originally Posted by Induction View Post
    I'd like to ask another question 'cause I like your approach. If every letter was repeated; what would happen? 28! would be divided out by the permutations of the repeated letters? Or is it still 27! divided out by the permutations of the repeated letters?
    If all the letters are the same then there is only one way to seat them.
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    Re: Permutation problems (circular & with repetition)

    "If every letter was repeated; what would happen?".

    I mean if A, B, C, D, E, F, G and H all have repeats (there's at least two of every letter); what happens with that?

    Looking back at what I wrote; it can't be 28! or 27! (for the numerator)... since I'm increasing the total number of letters (making repeats for each letter). This should fix some ambiguity with my question.
    Last edited by Induction; Oct 11th 2017 at 07:04 AM.
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    Re: Permutation problems (circular & with repetition)

    Quote Originally Posted by Induction View Post
    "If every letter was repeated; what would happen?".

    I mean if A, B, C, D, E, F, G and H all have repeats (there's at least two of every letter); what happens with that?

    Looking back at what I wrote; it can't be 28! or 27! (for the numerator)... since I'm increasing the total number of letters (making repeats for each letter). This should fix some ambiguity with my question.
    Suppose there are $a \cdot A, b\cdot B, c\cdot C, d\cdot D, e\cdot E, f\cdot F, g\cdot G, h\cdot H$. Here, I am using the notation for multiplicity. $a\cdot A$ means the number of available letter $A$'s we have is $a$. So, $a$ is a number while $A$ is a letter.
    Fix the position of one of the H's. Now, there are:

    $\dfrac{(a+b+c+d+e+f+g+h-1)!}{a!b!c!d!e!f!g!(h-1)!}$

    ways to permute the remaining letters.
    Last edited by SlipEternal; Oct 11th 2017 at 07:29 AM.
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    Re: Permutation problems (circular & with repetition)

    Quote Originally Posted by SlipEternal View Post
    Suppose there are $a \cdot A, b\cdot B, c\cdot C, d\cdot D, e\cdot E, f\cdot F, g\cdot G, h\cdot H$. Here, I am using the notation for multiplicity. $a\cdot A$ means the number of available letter $A$'s we have is $a$. So, $a$ is a number while $A$ is a letter.
    Fix the position of one of the H's. Now, there are:

    $\dfrac{(a+b+c+d+e+f+g+h-1)!}{a!b!c!d!e!f!g!(h-1)!}$

    ways to permute the remaining letters.
    I am mistaken. I apologize. I am assuming that the multiplicity for $H$ is 1. If the multiplicity for every letter is greater than 1, this become trickier, as there becomes potentially more symmetries.
    Thanks from Induction
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