True, but in general, repeat solutions can not be so easily eliminated. So, if someone can put forward a logical position for dropping a permutation or class of permutation in the analysis, this is permitted in the contest? Validation of the resultants could be difficult.
Jooel,
Parenthesis permutations:
(n n n ) n n (n n n) (n n) n n n (n n) n n n (n n) (n n) (n n)
I got 128 operator permutations by:
2*4*4*4
but Jens' point reduces this to 4*4*4=64
thus my revised number is: 9,216.
Of course there will be many solutions that are effectively repeat solutions in these permutations. Maybe we could might try to identify methods that eliminate repeats?
Note, Jens, Commonsense says you are correct but can you put forward a proof of this point covering all permutations. The presence of divisions which are not transative and parenthesis makes this a more difficult proof than I orginally thought.
[ 15. September 2002, 20:20: Message edited by: Jack Lothian ]
