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I'd like to confirm, what is the name of this combination, and if its calculated right:

Set (a,b) is spread over set of (1,2,3)

1 2 3 a b  a1 b1 2 3 a1 b2 3 a1 2 b3 b1 a2 3 b1 2 a3 1 a2 b2 3 1 a2 b3 1 b2 a3 1 2 a3 b3 

2*3 + 3 = 9

And based on that I'm doing it with a bigger sets without manual trial and error:

1 2 3 4 5 6 7 8 9 10 11 12 a b c d e f g 

7*12 + 12 = 96

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    The meaning is unclear. Your worked example suggests you are counting the maps from {a,b} to {1,2,3}, which is 3^2 = 9. However in the second case the answer would be much bigger than 96 if that were the meaning.2012-03-25

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If $a$ can take any of $3$ possible values, and independently $b$ can take any of $3$ possible values, then in total there are $3 \times 3 =9$ possibilities.

So with $m$ letters each taking $n$ possible values there are $n^m$ possibilities.

With $7$ letters and $12$ values this is $12^7 = 35831808$ possibilities, rather more than your $96$.

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    @PHPGAE: Try with one letter and twelve values: the answer is obvious $12$ possibilities. Now with a second letter, that has $12$ possible values too so combined you have $12 \times 12$ possibilities. Now with a third letter, that has $12$ possible values as well so combined you have $12 \times 12 \times 12$ ...2012-03-25
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If I understood your question correctly general formula should be :

$N= \frac{n!}{(n-k)!}+n$

where $n$ is a number of elements of numbers set , and $k$ is number of elements of letters set .

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    Oh, I must have messed with my mac calculator cause I got undefined number with calculation, but yes 0! gives 1 now. I can't e$x$plain much more, but I just want to get all positions letters can take on number set. Say there are 6 different sections on a circle plate and 3 different items to put on sections. If you throw items on a plate, they take random places and I want to get the count of combinations they (3 three items on 6 sections) can take.2012-03-26
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Actually @Henry might be right after all. I made further construction with a set of 1, 2, 3, 4 and a, b:

a1 b1 2 3 4 a1 b2 3 4 a1 2 b3 4 a1 2 3 b4 b1 a2 3 4 b1 2 a3 4 b1 2 3 a4 1 a2 b2 3 4 1 a2 b3 4 1 a2 3 b4 1 b2 a3 4 1 b2 3 a4 1 2 a3 b3 4 1 2 a3 b4 1 2 b3 a4 1 2 3 a4 b4 

To compare my earlier formula that coincidently happened to give right answer, we can see:

2*4 + 4 = 12 != 4^2 = 16

I was confused with the pattern, that you can see from the example result. There are 4 rows popping up from other rows with certain interval. I think that pattern kind of "fractalizes" depending how many items there are on sets. But thats another story.