##### Contents

Combinatorics

Probability & Statistics

# 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
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!

```

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