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 March 23rd, 2014, 08:08 PM #1 Newbie   Joined: Mar 2014 Posts: 3 Thanks: 0 Number of Combinations of Variable Percent Items Hello guys, hopefully you can help me with this, because I have no idea how'd you solve this problem. First Part: So I have 21 flavors and I want to figure out the total number of possible combinations. Meaning all possible combinations period, whether you choose 2 flavors, 3 flavors, choose 16, or combined all 21 flavors, etc. Second Part: Lets take this a step further though. To my knowledge, the number that results from the first part is only accounting for flavors that are mixed proportionally. Meaning if you mixed flavor A with Flavor B, the result would be a 50/50 mix of flavor A and flavor B. If you mixed flavor A, B, and C, the result would be a 33.3/33.3/33.3 mix of flavors A, B and C. So on and so forth. What I really want to know is the total number of combinations you can get if you also include all the combinations that are unevenly mixed. For instance if you were to imagine that every flavor can each be broken down into a 100% bracket. Meaning you could make a combination that is 60% A and 40% B. Or 99% A and 1% B. Or 1% A and 99% B. Therefore just the combination of the first two flavors could result in hundreds if not thousands of combinations. And then when you factor in the fact there is 21 flavors, each of which can be broken down into a 100% bracket, and you could make combinations using all flavors or just 2, or 16, or 7, or 12, or wtvr. I'm guessing the answer is an unfathomably large number, but I'm wondering how you would even go about calculating that and what sort of custom equation you would have to use. Any help would be great. Hopefully my explanation is thorough enough for you guys to understand what I'm getting at. Thanks!
 March 23rd, 2014, 08:48 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Number of Combinations of Variable Percent Items I don't even want to think about this one.... but each 1% of each flavour becomes a "flavour", so you have 2100 flavours....g'nite....
 March 23rd, 2014, 08:52 PM #3 Newbie   Joined: Mar 2014 Posts: 3 Thanks: 0 Re: Number of Combinations of Variable Percent Items Okay so you would just calculate all the possibilities as if you had 21 thousand flavors? What is the answer then?
March 23rd, 2014, 08:53 PM   #4
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Re: Number of Combinations of Variable Percent Items

Quote:
 Originally Posted by Damios Okay so you would just calculate all the possibilities as if you had 21 thousand flavors? What is the answer then?
Sorry that was meant to be 2100 flavors.

 March 24th, 2014, 10:11 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Number of Combinations of Variable Percent Items Well IF it is 2100: 2^2100 = 145.....376 : total of 633 digits!

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