# Permutation Enumeration

How
many different arrangements of letters

**a, b**, and**c**are formed? Through direct enumeration we get 6, namely**abc, acb, bac, bca, cab, and cba**. Each of these arrangements is called permutation. So there are 6 permutations that may be formed from 3 objects. This result can also be obtained from the principle above. Because the first letter in the permutation can be any of 3 letters, the second letter in this permutation can be selected anywhere from the remaining 2 letters, and the third letter is taken from the remaining 1 letter. So all of them have 3 x 2 x 1 = 6 possible permutations.
Suppose
we now have n objects. Similar reasoning as in the previous paragraph shows
that there is

*n (n-1) (n-2) ... 3x2x1 = n!*
permutations
that are different from n objects

**Example of Permutation 1**

How
many turn sequences hit batting for a baseball team consisting of 9 players?

Solution:
If using the formula

*n (n-1) (n-2) ... 3x2x1 = n!*Then the result is 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362 88. different order.**Example of Permutation 2**

An
opportunity theory class consists of 6 men and 4 women. An exam is given and
then students are sorted according to the results of their exam scores. Suppose
there are no two students who get the same score.

1. How
many ranks are possible?

2. If
men are sorted among themselves and women are also sorted among themselves, how
many ranks are possible?

Completion of the question answer:

1.
10! = 3 628 800

2.
(6!) (4!) = 17 280

**Examples of Permutation Problems 3**

Jhon
has 10 books that he wants to compose on his bookshelf. Among the books, 4 are
math books, 3 chemistry books, 2 history books, and 1 language book. Jhon
wanted to arrange his books so that all similar books were arranged not
separately on the bookshelf earlier. How many possibilities are different
arrangements?

Solution:
There are

**4! 3! 2! 1!**composition with math books in the first place, then chemistry books, then history books, and finally language books. Likewise, for each type of book, there are**4! 3! 2! 1!**possible arrangement. So, because there are 4! order of possible types of books, then the desired answer is**4! 4! 3! 2! 1! = 6912**.
Now
we will determine the number of permutations of objects if a certain number of
objects cannot be distinguished from one another. To clarify what is meant,
refer to the example Permutation problem 4 below,

**Examples of Permutation Problems 4**

How
many different letters can be formed from the letters P E P P E R?

Solution:
First note that there are 6! permutation of the letters P1 E1 P2 P3 E2 R if all
three P and both E can be distinguished from one another. However, pay
attention to one of these permutations, P1 P2 E1 P3 E2 R. If we now permute the
three letters P and also the two letters E among themselves, then the
arrangement they produce remains in the form of PPEPE R. That is , all 3! 2!
Permutation

P1
P2 E1 P3 E2 R P1 P2 E2 P3 E1 R

P1
P3 E1 P2 E2 R P1 P3 E2 P2 E1 R

P2
P1 E1 P3 E2 R P2 P1 E2 P3 E1 R

P2
P3 E1 P1 E2 R P2 P3 E2 P1 E1 R

P3
P1 E1 P2 E2 R P3 P1 E2 P2 E1 R

P3
P2 E1 P1 E2 R P3 P2 E2 P1 E1 R

Shaped
PPEPER. so, there are 6! / 3! = 60 possible arrangements of PPEPER letters.

In
general, similar reasoning used in showing that there is

permutations
that are different from n objects, of which n1 are the same, n2 of them are the
same, ... n, of which are the same

**Example of Permutation 5**

In
how many ways can 3 white marbles, 4 red marbles, and 4 black marbles be
arranged in a row when sam colored marbles cannot be distinguished from one
another?

Solution:
There are 11! / 3! 4! 4! = 11 550 possible arrangements

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