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Are there different ways to arrange all combinations without repeats other than the following method.

number of items = 6| group size = 4

1,2,3,4 | 1,2,3,5 | 1,2,3,6

1,3,4,5 | 1,3,4,6

1,4,5,6

2,3,4,5 | 2,3,4,6

2,4,5,6

3,4,5,6

James Beezho
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2 Answers2

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You're missing $(1,2,4,5), (1,2,4,6), (1,2,5,6), (1,3,5,6), (2,3,5,6)$ . These along with your combinations make $15 = {6 \choose 4}$ combinations, so these are all the possibilities.

As far as different methods go, that is actually the most efficient way.

William Stagner
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  • Thank you for this input. I had been doing it wrong for some time. Do you know of a way that tries to spread evenly each element of the elements to be combined along a vector of all combinations, so that 1 is equally likely to be in the first, second, third, to last possible combinations – James Beezho Apr 09 '15 at 18:06
  • Well, you could take the "matrix" of combinations formed by your method and look at the transpose. – William Stagner Apr 09 '15 at 20:45
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Yes! Think of it this way: there are 6 choices for the first number, 5 for the second, 4 for the third, and 3 for the fourth. 6*5*4*3=360

btw if the order matters, which I believe it does in the way you've stated the problem, you are not looking for combinations, but instead permutations.

finagle29
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