Das ist ein Modul aus dem "Defects" Hyperscript, der aber hier auch sehr gut paßt.  
Here we just look at the different ways to generate combinations or variations of "things" (= elements) belonging to a certain set of things.  
The set of "things" could be the numbers {0, 1, 2, ..., 9}; the letters of the alphabet; the atoms of a crystal; the people on this earth, in Europe, or in your hometown  you get the drift. We generally assume that the complete set has N such elements. The elements can be different or identical  the set {1,2,3,4,5 } has 5 different elements, the set {1,1,1,1,1} has 5 identical elements, for example.  
We then define subsets {k} that contain k elements from the set {0...9}; eight letters, a certain number n of atoms, and ask what kind of combinations or variations are possible between N and k.  
Note that we do not primariliy ask what you can do with k elements after you made a choice.  
To make that clear: If we have, for example, N = {0, 1, 2, ..., 9}, and k = 3, we may ask: How many possibilites are there to pick three members of N? That's a "legal" question. However, we do not first ask: How many different numbers can I form with the subset {2,4,5}}?  
That seems to be a good question, so why don't we allow it? Because the set {k} = {2,4,5} has no relation anymore with {N}. How many numbers you can form with the integers 2,4,5 is completely independent of {N}; so it is not an eligible question if want to look at relations between {N} and {k}. This does not mean that we can not answer the question of how many numbers..., too  if we phrase it correctly, that is!  
This is a bit abstract, so let's look at examples:  
For the first example we may ask:


It is obvious: Even for the most simple examples, there is no end of questions you can ask concerning possible arrangements of your elements.  
Some answers to possible questions are rather obvious, some certainly are not. For some, you might feel that given enough time you would eventually find the answer; some you might feel are hopeless  just for you, or possibly for everybody?  
Moreover, for some answers you have a feeling or some rough idea of what the result could be. It's just clear that all problems involving three digits have less than 1.000 possibilities, and that with more restrictions the number of possibilities will decrease. For other problems, however, you may not have the faintest idea of what the result might be. That is a big problem and makes combinatorics often very abstract.  
How to be systematic about this? That is an easy question: Study combinatorics  a mathematical discipline  for quite some time and you will find out.  
In particular you will find out that there is a small number of standard cases that include many of the typical questions we posed above, and that there are standard formulas for the answers. Let's summarize these standard cases in what follows. 
Quite generally, we look at a situation where we have N elements and ask for the number of arrangements we can produce with k of those elements.  
Some Examples:


The questions we ask, however, are not yet specific enough to elicite a definite answer. We have to construct 2 × 2 = 4 general cases or groups of questions.  
First we have
to distinguish between two basic possibilities of selecting elements for the
combinatorial task:


Second, we
have to distinguish between possibilities of
arranging the elements. An
arrangement in this sense, simply speaking,
can be anything that allows to visualize the combinations we make with the
elements selected  e.g. a string as shown above. We than have two basic
possibilities:


Sticking to natural numbers as elements of the set {N} for examples, we now can produce the following table for the four basic cases:  


Since the fraction marked in red comes up all the time in combinatorics, it has been given its own symbol and name.  
We define the binomial coefficient of N and k as  


Yes  it is a bit mind boggling. But it is not quite as bad as it appears.  
The first column gives an obvious result. How many three digit numbers can you produce if you have 0  9 and every possible combination is allowed (i.e,. 001 = 1 etc.) and counted. Yes  all numbers from 000, 001, 002, ..., 998, 999  makes exactly 1000 combinations, or C^{D}(k, N) = 10^{3} as the formula asserts.  
Always ask yourself: Am I considering a variation (all possible arrangement counts) or a combination ("indistinguishable" ^{1)} arrangements don't count separately)?  
Look at it from the practical point of view, not from the formal one, and you will get into the right direction without too much trouble.  
The rest you have to take on faith, or you really must apply yourself to combinatorics.  
All more complicated questions not yet contained in the cases above  e.g. we do not allow the element "0" as the first digit, we allow one element to be picked k_{1} times, a second one k_{2} times and so on, may be constructed by various combinations of the 4 cases (and note that I don't say "easily constructed").  
Arrangement of Vacancies
OK. For the example given the cases may be halfway transparent. But how about the arrangement of vacancies in a crystal? What are the elements of this combinatorial problem, and what is k?  
The elements obviously are the N atoms of the crystal. The subset k equally obviously selects k = n_{V} = number of vancanies of these elements.  
Now what exactly is the question to ask? There are often many ways in stating the same problem, but one way might be better than others in order to see the structure of the problem.  
We could ask for example:


It is clear now that we have to take the "identical elements" and "different arrangements of the same elements do not count" case  which indeed gives us the correct formula that we derived from scratch in exercise 2.14  
^{1)} There is a certain paradoxon here: In order to explain in words that certain arrangements are indistinguisghable, we have to list them separately, i.e. we distinguish them. But that is not a real problem, just a problem with words.
5.3.3 Gleichgewichtskonzentration von atomaren Fehlstellen in Kristallen
5.3.2 Definition der Entropie und erste Anwendung
Ableitung der Gauss Verteilung
Übung 5.32 Gleichgewichtskonzentration an Leerstellen
Übung 51 Anordnungsmöglichkeiten
© H. Föll (MaWi 1 Skript)