combinatorial analysis

Permutations

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

Question:
How many numbers of three digits can we obtain from 
this set?. The answer is:
123, 132, 213, 231, 312, and 321. In total, 6 numbers.

The order of these numbers is NOT important. The arragement 
132 and 312 , for example, are not the same.

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

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

We process as we did for the arragement. 
We arrange n objects among n objects. 
Here the arrangement is called permutation:

Since we know the number of counts for arragements, we induce 
the one for permutations, simply by setting r = n. It 
follows:


The number of permutations of n objects is 
P(n) = A(n,n) = n!