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So let's say we have $a,b,c,d$. In how many ways we can combine them in groups of 3. For example: $abc,acb,bca,aab,aac,bbc,ddc,ddd,aaa$. There should be a formula using arrangements. I did it long ago and I got like 64 if I remember well. By the ways the numbers can repeat 2 times or 3 times ex.:$aab,ccd,ccc$.

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    $\begingroup$ The number of ways to make three independent choices of 1 letter out of 4 is $4 \cdot 4 \cdot 4$. $\endgroup$
    – Umberto P.
    Feb 27 '17 at 15:37
  • $\begingroup$ So It is 64? Are you sure? Also can you tell me why it is like that? $\endgroup$
    – Ghost
    Feb 27 '17 at 15:37
  • $\begingroup$ Yes, and I explained why if you carefully read the comment. $\endgroup$
    – Umberto P.
    Feb 27 '17 at 15:38
  • $\begingroup$ emathematics.net/combinavrepeticion.php $\endgroup$ Feb 27 '17 at 15:40
  • $\begingroup$ Ok but I want a math explanation. $\endgroup$
    – Ghost
    Feb 27 '17 at 15:41
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I think you are looking for permutations with repetition, where in your particular case, the set is $S=\{a,b,c,d\}$, i.e. is a $(K=4)$-element set, and you are forming $(n=3)$-tuples.

The answer is, therefore $K^n=4^3.$

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