On 1 May 2001 15:43:50 -0700, [EMAIL PROTECTED] (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
>
>
>
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It could be I'm not understanding your colleague's problem, but if
one is looking for "repeats" of let us say 3 letters rather than
"1,2,3", then there are 27 such sets ---not 10:
AAA AAB AAC ABA ABB ABC ACA ACB ACC BAA BAB BAC BBA BBB BBC BCA BCB
BCC CAA CAB CAC CBA CBB CBC CCA CCB CCC
It seems to me then that 3^3 or 27 is the no. of "repeats." If there
are 4 letters then 4^4 or a total of 256 sets of "repeats" are found.
Given 3 items taken 3 at time provides 1 combination and 6
permutations I would think.
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