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# 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?

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