Formula Table

Ordered Repetitions Unique Definition Example Allegory
- - - $C_{n}^{k} = \frac{n!}{k!(n-k)!}$

\begin{eqnarray*} S &=& \{C_{n}^{3}|n=3,k=2\} \\ &=& \{ (1,2), (1,3), (2,3)\} \end{eqnarray*}

, $|S| = 3$

Combinations
x - - $A_{n}^{k} = \frac{n!}{(n-k)!}$

\begin{eqnarray*} S &=& \{A_{n}^{k}|n=3,k=2\} \\ &=& \{ (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) \} \end{eqnarray*}

, $|S| = 6$

Arrangements
x x - $n^{k}$

\begin{eqnarray*} S &=& \{n^{k}|n=3,k=2\} \\ &=& \{ (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) \} \end{eqnarray*}

, $|S| = 9$

Permutations with Repetitions
Term Usage
Combinations Food items in a salad that can only be selected together with other items.
Arrangements Sorting a subset of items in order such that each item can only be selected at most once.
Permutation A special case of "Arrangements" where the items to select are equal to the selected group size.
Permutations with Repetitions A combination lock.

Note that classical permutations $P_{n}$ can be seen as a special case of arrangements when $A_{n}^{n}$ or $A_{n}^{n-1}$ such that $A_{n}^{n}=A_{n}^{n-1}=P_{n}=n!$ due to $A_{n}^{n}=A_{n}^{n-1}$. In that sense, arrangements can be seen as a generalization of permutations when only some subset of the available items (the items to chose) have to selected.

Having said that, an interesting "brain-bug" is to immediately think about "permutations", as in $P_{n}$, as suitable to calculate the number of settings on a combination lock or a padlock when, in fact, "permutations with repetitions" is the correct response.

To illustrate the difference between both $P_{n}$ and $n^{k}$ cases by using a combination lock the following distinction can be made: