Posted by Denis Borris on October 22, 2002 at 13:36:29:
In Reply to: 74,005,152 for "35" confirmed... posted by Denis Borris on October 22, 2002 at 02:05:32:
...finally figured out the sumbitch (probably a formula everybody knows about):
Take 1 1 1 3 3 3 3 3 3 4 5 5 (one of 197 "number combos" that add up to 35)
We have 3 of same, 6 of same, 1 of same, 2 of same:
12! / (3! * 6! * 1! * 2!) = 55440
n! / [n(1)! * n(2)! ... * n(i)!]
Is this an acceptable "way" to show the general case?
Proper programming using that formula should produce results for
all of the 12 to 72 possibilities in a jiff....
Right, Joel :)