🚀 go-pugleaf

Andrew wrote:
> Anyone have any pascal code for all possible combinations of a set?
> 
> So if we start with '1','2','3' the possible combinations are:
> 
> 1
> 2
> 3
> 1 2
> 1 3
> 2 3
> 1 2 3


number of combinations

= number of combinations 1 of 3
  + number of combinations 2 of 3
  + number of combinations 3 of 3

    3       3       3
= (   ) + (   ) + (   )
    1       2       3

= 3+3+1

= 7
       a
with (   ) = a! / (b! (b-a)!)
       b

I think you missed the combination 0 of 3 in your list. Unless this
was                             3
intended, you have to add (   ) = 1. Then the result would be 8.
                            0

                         n
                        ---
                        \     n
generalized: num of c =  >  (   )  for a set with n elements.
                        /     k
                        ---
                        k=0

Translating this into Pascal is up to you.

Wolf