Multiply the number allowed in each group by the number allowed in the other groups. In your case 4*3*5 = 60. This is actually fun because you can figure out combos for everything. Like the lotto. If you have 60 balls in Lotto, and 6 balls are drawn, then the possible combos are 60*59*58*57*56*55 = 36,045,979,200. Therefore your odd are 1 in 36,045,979,200 in winning. Now if you could throw the ball back into the mix each time, then the odds skyrocket because then your possible combos are 60^6 which is 1 in 46,656,000,000. Sorry. I used to really suck at math but enjoyed doing it.
Michael Corrigan Programmer Endora Digital Solutions 1900 S. Highland Avenue, Suite 200 Lombard, IL 60148 630-627-5055 x-136 630/627-5255 Fax ----- Original Message ----- From: Phillip Broussard To: CF-Community Sent: Friday, February 15, 2002 9:29 AM Subject: Anyone good with math? I have a math problem and have no idea how to get the answer. Let's say that I have 3 groups. The first one has 4 objects, the second has 3 objects and the third has 5 objects. How would I figure out the total possible combinations I could have of the three groups if you could only have one from each group using coldfusion? Phillip Broussard Tracker Marine Group 417-873-5957 ______________________________________________________________________ Macromedia ColdFusion 5 Training from the Source Step by Step ColdFusion http://www.amazon.com/exec/obidos/ASIN/0201758474/houseoffusion Archives: http://www.mail-archive.com/[email protected]/ Unsubscribe: http://www.houseoffusion.com/index.cfm?sidebar=lists
