Formulas:

1. Arrangements:

We can arrange a set with or witout repetition:

Set = {1,2,3,4,5}, with N elements.

Set arranged by three elements without repetition:
Set3 = {1,2,3} or {2,3,1}, or {3,2,5} , ...

Set arranged by three elements with repetition:
Set3 = {1,2,2} or {3,5,5} , ...

The subsets {1,2,3} and {2,3,1} are differents because 
the order is important

1. without repetition:

We have N possibilities for the first choice, (N - 1) possibilities for the 
second choice, ..., and (N - (n - 1)) = (N - n + 1) possibilities for the last 
nth choice. In total: 
N x (N - 1) x (N - 2) x ... x (N - n + 1) = the same factor x new_term/new_term, 
where new_term = (N - n) x (N - n - 1) x ... x 2 x 1 = (N - n)!
We have the same factor x new_term = N!
Finally:

Without repetition, the number of ways to arrange n elements 
chosen among N is A(n,N) = N!/(N - n)!


2. with repetition:

We have N possibilities for the first choice, N possibilities for the 
second choice, ..., and N  possibilities for the last nth choice. In total: 
N x (N x ... x N (n times) = Nn
Finally:

With repetition, the number of ways to arrange n elements 
chosen among N is A(n,N) = Nn

2. Combinations:

Set = {1,2,3,4,5}

Set of three combined elements without repetition:
Set3 = {1,2,3} but NOT {2,3,1}, and NOT {1,1,2}, or  ...

Set of three combined elements with repetition:
Set3 = {1,2,2} or {3,5,5} , ...

The subsets {1,2,3} and {2,3,1} are equal because the order is NOT important


The number of ways to combine, without repetition, n elements 
chosen among N is C(n,N) = A(n,N)/n! = N!/n!(N - n)!



The number of ways to combine, with repetition, n elements 
chosen among N is C(n,N) =  (n + N - 1)!/n!(N - 1)!

3. Permuations:

Set = {a,b,c}

The all sets of permuted elements without repetition:
{a,b,c}, {a,c,b}, {b,a,c}, {b,c,a}, {c,a,b}, {c,b,a}

All the  sets are different because the order 
is important. Permute a set is just arrange all of its 
elements; that is arrage N elements among N, or 
A(N,N) = N!/(N - N)! = N!


The number of ways to permute N elements is  P(N) = N!


Remarks:
1. An arrangement is a permutation, we have just to 
specify the numbre of elements. 
If n &nequi; N ⇒ arrangement, and if n &equi; 
N ⇒ permutation.

2. A combination is an arrangement when the order does 
not matter.
If we arrange, without repetition, n objects among N, 
we have A(n,N) ways. Each time we arrange n objects, they 
are ordered n! times. Then the number of ways to combine n 
elements, without repetition, among N is 
C(n,N) = A(n,N)/n! = N!/n!(N - n)!.