Dale Glaser wrote:
> Hi there..I have scoured my admittedly limited collection of probability
> texts, and am stumped to find the answer to the following, so any help is
> most appreciated....a colleague just approached me with the following
> problem at work: he wants to know the number of possible combinations of
> boxes, with repeats being viable.......so, e.g,. if there are 3 boxes then
> what he wants to get at is the following answer (i.e, c = 10):
>
> 111
> 222
> 333
> 112
> 221
> 332
> 113
> 223
> 331
> 123
>
> ...so there are 10 possible combinations (not permutations, since 331 =
> 133)...however, when I started playing around with various
> combinations/factorial equations, I realized that there really isn't a pool
> of 3 boxes..there has to be a pool of 9 numbers, in order to arrive at
> combinations such as 111 or 333.............so any assistance would be most
> appreciated as I can't seem to find an algorithm in any of my
> texts..........thank you.........dale glaser
maybe we need some more conditions on the problem? too many options here.
Suppose there are 3 unique boxes. How many ways can I stack the 3 of them?
Answer: If I count permutations, it would be 3P3 (read sub scripted numbers
here), or 3!/(3-3)!= 3! = 6
But you said not to count permutations. For combinations: 3C3 =
3!/[0!*3!] =
1 Silly of course.
Suppose you can re-use a box, so 222 is OK. first position has 3 possibles,
same with 2nd pos, and 3rd.
Total: 3*3*3 = 27 If you set them out in a careful sequence, it will become
more evident:
1 1 1
2 1 1
3 1 1
1 2 1
2 2 1
3 2 1
1 3 1
2 3 1
2 2 1
etc.
--
Jay Warner
Principal Scientist
Warner Consulting, Inc.
4444 North Green Bay Road
Racine, WI 53404-1216
USA
Ph: (262) 634-9100
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