number of elements in unsortet case

Question

I have a group M with Mn different elements. How many unique combinations can I make out of this in an n digit system when order is no importance.

For example if M = {1 2} & n = 3

1 1 1
1 1 2
1 2 2
2 2 2

Answer 1

Just to be clear: You have a set $M$ consisting of $m$ distinct members, from which you wish to make $n$ selections, with repetition, where the order of selection is indistinct. Correct?

Use Stars and Bars.

This is essentially the same way to assign $n$ indistinct stars to $m$ distinct bins, where the bins are the $m$ elements in the set $M$, and the stars count the number of times each is selected, for a total of $n$ selections. The ways to arrange this is:

$${n+m-1 \choose n} = {n+m-1 \choose m-1}$$

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So as in your example, when $m=2, n=3$ that is: $\frac{(n+m-1)!}{n!\;(m-1)!} = \frac{4!}{3!\;1!} = 4$