Wikipedia:Reference desk/Archives/Mathematics/2009 January 1

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January 1 edit

Simple permutations/combinations? edit

I'm sure this is really simple, but for some reason I can't figure it. Anyway...

I have two jars, A and B. I care only which balls it contains, not the order of the balls or anything. So let's say there are two balls, with numbers written on them (1 and 2). There are 4 combinations in this case, A1 B2, A2 B1, A12 B, A B12. If I add a 3rd ball, there are now 8 combinations (A31 B2, A32 B1, A312 B, A3 B12, A1 B32, A2 B31, A12 B3, A B312). If I add a 4th ball, there are now 16 combinations (can't be bothered to list them all). I'm guessing if I add a 5th ball, there will be 32, a 6th ball, 64, etc.

What I'm looking for is an equation to show the relationship between n (number of balls) and t (total combinations). I can't figure it out but it was in a chemistry text book that I no longer own. --Mark PEA (talk) 21:40, 1 January 2009 (UTC)[reply]

You seem to have already guessed the relationship. What's the problem? Algebraist 21:45, 1 January 2009 (UTC)[reply]
Oh my god.. I can't believe all the time I've wasted. Of course it is t = 2^n. Maybe I'm deprived of sleep, can't believe I didn't see that straight away. --Mark PEA (talk) 21:48, 1 January 2009 (UTC)[reply]
This is your god replying. How about the general formula for any number j jars into which are put n balls? Cuddlyable3 (talk) 19:07, 2 January 2009 (UTC)[reply]
A pure guess would be t = j^n. Is it? --Mark PEA (talk) 00:04, 4 January 2009 (UTC)[reply]
Prove it! Algebraist 00:05, 4 January 2009 (UTC)[reply]
Know ye that a god can take away your balls leaving you with n = 0. Cuddlyable3 (talk) 14:54, 6 January 2009 (UTC)[reply]