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BSvK
science forum beginner
Joined: 02 Oct 2005
Posts: 1
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 Posted: Sun Oct 02, 2005 5:01 pm Post subject:
counting words
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Consider four sets of symbols, first of m alphas, second of m betas,
third of n gammas, fourth of n deltas. These are to be arranged in
2(m+n)-long words such that no pair of identical symbols are adjacent. I
would like to count the number of different words so formed. I think Polya
theory should apply here but I was unable to do so. Please, help. |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
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| Posted: Tue Oct 04, 2005 4:33 am Post subject:
Re: counting words
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BSvK wrote:
| Quote: | Consider four sets of symbols, first of m alphas, second of m betas,
third of n gammas, fourth of n deltas. These are to be arranged in
2(m+n)-long words such that no pair of identical symbols are adjacent. I
would like to count the number of different words so formed. I think Polya
theory should apply here but I was unable to do so. Please, help.
|
You need to supply the following information, unless you want the most
general solution:
How many alphas, betas, gammas, and deltas are there?
How many alphas are also betas?
How many betas are also gammas?
How many gammas are also deltas?
--- Christopher Heckman |
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Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
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| Posted: Wed Oct 05, 2005 6:24 am Post subject:
Re: counting words
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Proginoskes wrote:
| Quote: | BSvK wrote:
Consider four sets of symbols, first of m alphas, second of m betas,
third of n gammas, fourth of n deltas. These are to be arranged in
2(m+n)-long words such that no pair of identical symbols are adjacent. I
would like to count the number of different words so formed. I think Polya
theory should apply here but I was unable to do so. Please, help.
You need to supply the following information, unless you want the most
general solution:
How many alphas, betas, gammas, and deltas are there?
How many alphas are also betas?
How many betas are also gammas?
How many gammas are also deltas?
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BSvK has responded privately, saying:
| Quote: | There are m alphas, m betas, n gammas, n deltas. Alphas, betas ,gammas,
deltas are all different from each other.
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This is easy then. The number of ways to choose the first alpha is m,
the second way m-1, the third way m-1, etc., so the number of strings
of alphas with no consecutive identical symbols is m*(m-1)^(m-1). This
will also be the number of strings of betas. The number of gamma
strings will be n*(n-1)^(n-1), and this will also be the number of
delta strings. We can combine any alpha string with any beta string,
etc., so the total number of strings is
m*(m-1)^(m-1) * m*(m-1)^(m-1) * n*(n-1)^(n-1) * n*(n-1)^(n-1), or
m^2*(m-1)^(2m-2) * n^2 * (n-1)^(2n-2).
--- Christopher Heckman |
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