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August 17th, 2007, 04:20 PM  #1 
Newbie Joined: Aug 2007 From: Tennessee Posts: 1 Thanks: 0  Permutations Question
I hope I am posting this in the right place. I am not sure if I'm looking for combinations or permutations on this, so please bear with me. I have 3 different sauces that I can put on to one sandwich. The sauces are Ketchup, Mustard & Mayo. This is what I came up with (the goal is to have no sauces repeating themselves, so no single sauce put on twice). The sandwich has a top and bottom with the sauces in 15 different orders with anywhere from 1 to 3 sauces on it. I wrote out the problem on paper and drew little boxes with the possible solutions and these are the solutions I came up with... 1. Ketchup only 2. Mustard only 3. Mayo only 4. Ketchup & Mustard 5. Ketchup & Mayo 6. Mustard & Ketchup 7. Mustard & Mayo 8. Mayo & Ketchup 9. Mayo & Mustard 10. Ketchup, Mustard & Mayo 11. Ketchup, Mayo & Mustard 12. Mayo, Ketchup & Mustard 13. Mayo, Mustard & Ketchup 14. Mustard, Ketchup & Mayo 15. Mustard, Mayo & Ketchup I'm guessing these are the combinations. I still don't fully understand permutations. So, my question is, without having to draw the diagrams, how would I solve this? How would I know to multiply 3 by 5? Or do I even multiply by 5 and there's another process I don't know about? I appreciate any help on this. Thank you! 
August 17th, 2007, 04:44 PM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
Yep, that's right, there's 15 combinations. The reason is this: If all three sauces are put on: 3*2*1 combinations The reason is this: If you have three slots and 3 different sauces, then you could put on any sauce on at first, then any sauce except the first one, then any sauce except the first and second, etc. This leads to 3*2*1. If two sauces are put on: 3*2 combinations For the same reason. If only one sauce is put on: 3 combinations Pretty self evident, since we have only 3 types of sauces. Add them all up, and you get 15 combinations. NOTE: This assumes that you consider different orders of the sauces to be distinct ways of putting the sauces on the sandwich. For instance, this answer assumes that Mustard, Mayo, Ketchup is different than Ketchup, Mayo, Mustard. Otherwise, if you counted mirror reversals to be the same, there would only be 9 ways of putting the sauces on the sandwich, because each number of combinations involving more than two sauces would have to be divided in half. 
August 17th, 2007, 05:03 PM  #3 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Now, personally, I would consider mustard, ketchup, mayo to be the same as ketchup, mayo, mustard, but that's just me :P . If we assume that any combination of the same condiments is indistinguishable (as it probably will be when you bite into it), then we have three ways to select one condiment, three ways to select two (its the same as choosing which condiment to not put on your sandwich) and one way to choose all three, for a total of 7 combinations (of course, you could always just choose to not put any condiments on, which is another combination . This is a total of 8=2³ combinations, which is no coincidence.).


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