
#1
Jun804, 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? 



#2
Jun804, 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. 



#3
Jun804, 11:24 AM

P: 78

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



#4
Jun804, 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 indepth 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). 



#5
Jun804, 05:33 PM

Emeritus
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P: 11,154




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