how to find all the possible outcomes for these?

Question

lets say i have a group of number... 0,0,1,1,2,3
how can i find all possible 4 digit group from those number?? without the 4 digit group having a duplicate group... meaning that if i have 1 group consisting 0011, i dun need 0101, or 1100 or anything that comes out from this 4 digit together...
by the way... are there any program that is capable to do that??

Answer 1

In combinatorial mathematics, a combination is an un-ordered collection of unique elements. (An ordered collection is called a permutation.) Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in different orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once); this is often referred to as "without replacement/repetition". This is because combinations are defined by the elements contained in them, s the set {1, 1, 1} is the same as {1}. For example, from a 52-card deck any 5 cards can form a valid combination (a hand). The order of the cards doesn't matter and there can be no repetition of cards.

A k-combination (or k-subset) is a subset with k elements.

A permutation is an ordered list without repetitions, perhaps missing some (n−r) elements.

where:

http://s236.photobucket.com/albums/ff177/jsardi56/?action=view¤t=combinationsandpermutations10-31-07.jpg

In your case you are saying that 0011, 0101, and 1100 or any other arrangement of those four elements should be counted only once.
So you are counting combinations. For example if there are 6 numbres in your group and you want the number of sub groups containing 4 numbers, your calculation would be:
C(n, k) = n!/[k!(n - k)!] = 6!/[4!(6 - 4)! = 6x5x4x3x2/[(4x3x2)(2)] = 30/2 = 15

Example:
If n > 1 and r > 1, each permutations element order becomes significant. For example the permutations of 3 elements in the set {1, 2, 3} taken in sequence lengths of 2 is:

(1,2), (1,3), (2,1), (2,3), (3,1) and (3,2). 6 permutations for sequence lengths of 2
This should not be confused with combination where the order of the elements is not taken into account and the above elements in the same set combine into just the 3 combinations for the same lengths:

(1,2), (1,3) and (2,3). 3 combinations for collection lengths of 2