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Let's say I want to find a sample space of such experiment: There are three exits from the box: Left (L), right (R) and front (F). Three mice have been put in that box. Find the sample space of an event which mouse has chosen which exit. Function Permutations[] gives all different outputs.

L={L,R,F}
Permutations[L]

How can I find also outcomes like {F,F,L}, {R,R,R}... ?

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    $\begingroup$ It's only terminology. The OP has to state if he/she wants variations or combinations. (In other words: Is (LLF) the same as (LFL)?) $\endgroup$ – Stefan Feb 18 '13 at 14:48
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    $\begingroup$ @Ajasja When I use your example, I don't get the event {RRR} which the OP demanded. So Permutations does NOT give the asked results. $\endgroup$ – Stefan Feb 18 '13 at 15:02
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    $\begingroup$ @Misery Please try to re-state your question in an easier to understand way, to avoid all these close and reopen votes. I'd suggest giving an example input and the complete output your desire. $\endgroup$ – Szabolcs Feb 18 '13 at 15:10
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    $\begingroup$ Question should be restated as: Given a list of 3 distinct objects, how does one form the list of all possible triples of these objects, repetitions allowed? $\endgroup$ – murray Feb 18 '13 at 15:17
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    $\begingroup$ I disagree that his question is too localize: 4 upvotes and 6 upvotes on accepted answer shows that it is helping new visitors. I suggest removing the closure by lack of evidence supporting localization. $\endgroup$ – Jorge Leitao Oct 16 '13 at 9:32
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 Tuples[{l, r, f}, 3]
 (* {{l, l, l}, {l, l, r}, {l, l, f}, {l, r, l}, {l, r, r}, {l, r, f}, 
    {l, f, l}, {l, f, r}, {l, f, f}, {r, l, l}, {r, l, r}, {r, l, f}, 
    {r,  r, l}, {r, r, r}, {r, r, f}, {r, f, l}, {r, f, r}, {r, f, f},
    {f, l,  l}, {f, l, r}, {f, l, f}, {f, r, l}, {f, r, r}, {f, r, f}, 
    {f, f,  l}, {f, f, r}, {f, f, f}} *)

?

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The only problem is that you have a recursive definition ( you want L to be a list and outcome).

In general using identifiers that start with a capital letter is best avoided. D and C are some examples of this.

This is much safer and works as expected:

list = {"L", "R", "F"}
Permutations[list]

If you would like to have "permutations with repetition" then Tuples is the right tool for the job (as kguler already noted):

Tuples[list, 3]
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  • $\begingroup$ And E, and I, and N, and O... $\endgroup$ – Dr. belisarius Feb 18 '13 at 14:38

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