In this lesson, we will explore the concept of a permutation. A *permutation* is an arrangement or selection in which the order matters.

Consider this scenario: a college student has just purchased a new bookshelf for her dorm room and she has a total of *n* books that she wishes to arrange on the shelf. How many different ways can she arrange her books?

For the first spot on the shelf, she has *n* different options for the placement of a book. Next, since one book has already been placed, she now has *n-1* books to place in the second spot. She will then have *n-2* books for the third spot, *n-3* books for the fourth spot, and so on. Suppose she has nine books in total, the arrangement of the bookshelf will look like this:

`$\underline{9} \ \underline{8} \ \underline{7} \ \underline{6}\ \ldots \ \underline{1}$`

So there are:

`$9 \times 8 \times 7 \times \ldots \times 1 = 362,880$`

ways to arrange the books.

For a general arrangement of *n* items, we have:

`$n(n-1)(n-2) \ldots 1 = n!$`

Here, the *!* is the factorial operator.

Arranging *n* items without repetition results in *n!* a number of permutations.

Note the following identities:

- 0! = 1
- The factorial operation is undefined for numbers less than zero.

### Instructions

**1.**

Given the following set of letters:

`$Letters = \left \{ A, \ B, \ C, \ D \right \}$`

How many different four-character strings, without repetition, can be constructed out of this set?

Assign your answer to the `four_characters`

variable in the accompanying Python script.

**2.**

How many ways are there to arrange six chess pieces?

`$C = \left \{ pawn, \ knight, \ bishop, \ rook, \ queen, \ king \right \}$`

Assign your answer to the `chess_pieces`

variable.