set arrangements


by mikesvenson
Tags: arrangements
mikesvenson
mikesvenson is offline
#1
Jun8-04, 01:12 AM
P: 78
lets say I have a set of 2 numbers (1, and 2). I can arrange this set (including all possible amounts of numbers in the set) into 4 arrangements, 1, 2, 12, and 21.

For the set (1, 2, 3), I can arrange 15 different combinations of numbers contained in the set.

What about a set of 4 numbers? (1, 2, 3, 4)
What about a set of 5 numbers? (1, 2, 3, 4, 5)

Remeber I wish to count all possible combinations available within the set. e.g, 1, 2, 3, 4, 12, 21, 23, 32, 34, 43, 123, 321, 234, 432, 134, 431, .............1234, 1342, 1243, 4231, 4132, .........................

What is the formula for this?
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Muzza
Muzza is offline
#2
Jun8-04, 05:26 AM
P: 696
You want:

[tex]\sum_{k = 1}^{n} P(n, k) = \sum_{k = 1}^{n} \frac{n!}{(n - k)!}[/tex]

Maybe a "nicer" formula can be found.
mikesvenson
mikesvenson is offline
#3
Jun8-04, 11:24 AM
P: 78
Quote Quote by Muzza
You want:

[tex]\sum_{k = 1}^{n} P(n, k) = \sum_{k = 1}^{n} \frac{n!}{(n - k)!}[/tex]

Maybe a "nicer" formula can be found.
hmm, i hope so, i dont have a clue on the propor why do decifer this formula! What kind of math does that formula fall into? There must be a way to compute it without venturing into this unfamiliar territory!

I don't even know what the P's, n's, and k's stand for!

Muzza
Muzza is offline
#4
Jun8-04, 11:39 AM
P: 696

set arrangements


P(n, k) denotes the number of different permuations (make sure you know the difference between permutation and combination) of size k taken from a set of size n. See http://mathforum.org/dr.math/faq/faq.comb.perm.html for a more in-depth explanation. n stands for the number of elements in your set (I should've mentioned that).

Think about how you could calculate the number of "arrangements" that can be formed with elements from the set {1, 2, 3, 4}, without actually listing all of them. First you could count the number of "arrangements" with only 1 number. Then you count the number of "arrangements" with 2 numbers, and so on. Then you add all those numbers. But these "arrangements" are just permutations of size 1, 2, 3, or 4 from a set of size 4, and there's a function that can count these (namely P(n, k)). So the answer is P(4, 1) + P(4, 2) + P(4, 3) + P(4, 4). And as it turns out, P(n, k) is equal to n!/(n - k)!, so the answer is equal to 4!/(4 - 1)! + 4!/(4 - 2)! + 4!/(4 - 3)! + 4!/(4 - 4)!, which can "easily" be evaluated (it's 64).
Gokul43201
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#5
Jun8-04, 05:33 PM
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P: 11,154
Quote Quote by Muzza
You want:

[tex]\sum_{k = 1}^{n} P(n, k) = \sum_{k = 1}^{n} \frac{n!}{(n - k)!}[/tex]

Maybe a "nicer" formula can be found.
If the order was unimportant (ie. 12 is the same as 21) and the null set was included, then of course, this is just 2^n (cardinality of the power set), for a set of n elements...which is what you get by changing 'P' to 'C' in Muzza's expression and start summing at k=0.


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