Counting analysis

Arragements

Let's consider the following set of digits Set= {1,2,3}

Question:
How many numbers of two digits can we obtain from 
this set?. The answer is:
12, 13, 21, 23, 31, and 32. In total, 6 numbers.

The order of these numbers is NOT important. The arragement 
13 and 31 , for example, are not the same.

The question would be presented as: How many arrangements 
of two digits can we make from the three elements of the set Set?

There are 6 manners to arrange two digits among three digits.

In general, we process as follows, to arrange p objects among 
n objects:

There is n possibilities to choose the first object.
Once done, it remains (n-1) possibilities to choose the second object,
Once done, it remains (n-2) possibilities to choose the third object,
... etc ...
Once done, it remains (n-(r-1)) possibilities to choose the r th object.

In a total, there is A = n x (n-1) x (n-2) x ... x (n-r+1) arrangements 
of r objects among n objects. 

The result A can be written:
A = A x R/R 
where R = (n-r) x (n-r-1) x (n-r-2)... x 3 x x 2 x 1 
A x R = n!
R = (n-r)!
Finally:

The number of arrangements of r objects among n objects is  
A(r, n) = n!/(n-r)!