permutation
the arrangement of different objects into a linear order using each object exactly once is called a permutation of these objects. the number 
of all permutations of 
objects is called factorial, and is denoted by 
.
[cite:;taken from @combinatorics_bona_2023 definition 3.1]
in a permutation, order matters, repetition not allowed.
of all permutations of objects is called factorial, and is denoted by .[cite:;taken from @combinatorics_bona_2023 definition 3.1]
a permutation of a set of distinct objects is an ordered arrangement of these objects. we also are interested in ordered arrangements of some of the elements of a set. an ordered arrangement of 
elements of a set is called an -permutation.
[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations]
elements of a set is called an -permutation.[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations]
-permutations are also sometimes called variations.
if is a positive integer and is an integer with 
. then there are
-permutations of a set with distinct elements. note that when 
, we simply have
[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations; theorem 1]
is a positive integer and is an integer with . then there are-permutations of a set with distinct elements. note that when , we simply have[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations; theorem 1]if and are integers with 
, then 
.
[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations; corollary 1]
and are integers with , then .[cite:;taken from @discrete_kenneth_2018 chapter 6.3.2 permutations; corollary 1]
let 
be a permutation, and let 
. then there exists a positive integer 
so that 
.
[cite:;taken from @combinatorics_bona_2023 lemma 6.4]
be a permutation, and let . then there exists a positive integer so that .[cite:;taken from @combinatorics_bona_2023 lemma 6.4]
assuming there are 3 girls and 4 boys, we want them seated on a bench
such that the 3 girls are next to each other
if it wasnt for the condition that the girls must be seated next to each other we would have $7!$ permutations
since the girls are seated next to each other, its safe to look at the girls as a single element in a set of 5 such that each boy takes a seat and the fifth seat is reserved for a single girl, which means we have $5!$ permutations for this set
now since the girls can switch places and it wouldnt break the condition, because they'd still be seated one next to the other, this property has $3!$ permutations, so the total number of permutations for the seat is $5! \cdot 3!$
since the girls are seated next to each other, its safe to look at the girls as a single element in a set of 5 such that each boy takes a seat and the fifth seat is reserved for a single girl, which means we have $5!$ permutations for this set
now since the girls can switch places and it wouldnt break the condition, because they'd still be seated one next to the other, this property has $3!$ permutations, so the total number of permutations for the seat is $5! \cdot 3!$
such that the 4 boys sit all next to one another
the permutations of the boys sitting next to each other is $4! \cdot 4!$, because in total we'd have 3 girls and 1 boy (instead of 4), and the total permutations is $7!$ so the answer is $7! - 4! \cdot 4!$
permutation with repetition
given $n$ items and $k$ types, $n_1$ of the first type, $n_2$ of the second type and so on.. such that $n=n_1+n_2+\cdots+n_k$, then the total number of possible permutations with $n$ items is
circular permutation
when the permutations are circular, the number of options becomes $P_{n-1} = (n-1)!$