Posted by finelinebob on September 6, 2006, at 21:36:53
In reply to A math puzzle, posted by Alexus on September 6, 2006, at 1:37:34
There is a systematic way. It's called a Combination. Cominations can either allow for repetition or not, but since it doesn't matter what order the dx's are presented we'll do it without.
(in other words, if it mattered that you were BP I THEN ADD, and that was different than ADD THEN BP I, we'd need to include repetition).
General formula is n!/[k!(n-k)!. If you have 9 dx's and you need to find out how many combinations there are if you take them 5 at a time, then n=9, k=5, and the result would be 9!/[5! (9-5)!]
If you don't know what that "!" means, it's a factorial. 3! = 3*2*1 = 6. 0! is defined as 1.
Your problem is that you have more than one combination. You have these:
n=9, k=5
n=9, k=6
n=9, k=7
n=9, k=8
n=9, k=9
Add them all together and you get your answer.Oh. That would be 256.
http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/ComCount.htm#rjava3
flb
poster:finelinebob
thread:683578
URL: http://www.dr-bob.org/babble/social/20060901/msgs/683796.html