Combinations of two sets 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
 A: 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 $96$. 
A: 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 .
A: 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.
