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August 17th, 2007, 04:20 PM   #1
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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!
EternalSummer247 is offline  
August 17th, 2007, 04:44 PM   #2
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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.
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August 17th, 2007, 05:03 PM   #3
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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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