0
$\begingroup$

I want to find all the different combinations of dividing 17 into 2 groups; the order does not matter; each group has to contain at least one item.

I know the formula is C(17,r)=17!/(r!*(17-r)!) and my r is one to eight so there's a lot! I hoping there is an easy way to do this. I have been trying to assign items into the two groups by using 1 and 0 - i.e. 01111111111111111, 10111111111111111, 11011111111111111, ... 00111111111111111, 01011111111111111, etc ...

  • 0
    Excel has no direct function to present each partition. You may want to do this using another program and import the large result into Excel.2012-09-16

1 Answers 1

2

Well firstly there are 2^{17} subsets of the 17 objects (either an object is in or not in a subset). However 1 of these subsets contains no elements and another one contains all 17 of them (which would leave the other group having none). So these are discounted. So we are left with 2^{17}-2 ways of partitioning it. However we have double-counted as every subset we make has a complement that we count with this method but shouldn't count as the order of the 2 groups doesn't matter. therefore [2^{17}-2]/{2}=2^{16}-1 different combinations.