I thought I covered that in the second example by throwing the ball back into the mix after it was chosen which means that the number of combos is 60*60*60*60*60*60. I wrote it as 60^6 because that's how I had to notate it in the old piece off shit computers that the university I went to used for graphing these functions. I want to check out that book though. I was terrible at math when I was young, but I always enjoyed doing it. It wasn't until I was in my junior-senior year in high school that things started to click. Then I did much better at it in college. I help my nephews with their math homework and I think that I give them a good perspective because I know what they're going through.
BTW, he's in seventh grade and is doing algebra. Not basic algebra either. I was helping him last night figure out how to understand direct and inverse variations. I had to even look at the examples to remember how to do it. But it was fun though. Now if I can only get them to play chess!! 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: Simon Horwith To: CF-Community Sent: Friday, February 15, 2002 10:23 AM Subject: RE: Anyone good with math? of course, this changes when you have duplicate values in any of the groups, and duplicates are not allowed in your result set. Not too tough to figure out, but it makes things more interesting, for sure. By the way, for those of you interested in learning more about these typres of equations (probability and chance)... check out "Innumeracy: Mathematical Illiteracy and Its Consequences" (by John Allen Paulos). It's a very good book, and is a surprisingly easy read. ~Simon Simon Horwith Macromedia Certified Instructor Certified Advanced ColdFusion 5 Developer Fig Leaf Software 1400 16th St NW, # 500 Washington DC 20036 202.797.6570 (direct line) www.figleaf.com -----Original Message----- From: Michael Corrigan [mailto:[EMAIL PROTECTED]] Sent: Friday, February 15, 2002 10:43 AM To: CF-Community Subject: Re: Anyone good with math? 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
