# How do I understand permutations and combinations

## Combinatorics

### What is combinatorics?

The **Combinatorics** is a branch of mathematics and is sometimes assigned to statistics. In combinatorics one deals with possible arrangements of objects. More precisely, one deals with how many arrangements are possible.

Arrangements relating to all elements of a set are also made **Permutations** called. In the case of permutations, finds **no choice** instead of. Arrangements in which a selection is made can be found again in **Variations** and **Combinations** split up.

Where do you need such numbers?

For example, to calculate the probabilities of a **Laplace experiment**. These can be calculated with this formula:

$ P (E) = \ dfrac {| E |} {| \ Omega |} = \ dfrac {\ text {number of all f} \ ddot {\ text {u}} \ text {r} E \ text {g} \ ddot {\ text {u}} \ text {other results}} {\ text {number of all m} \ ddot {\ text {o}} \ text {similar results}} $

As you can see, you have to be able to determine numbers in this context.

### Permutations

The number of possible permutations depends on whether they are considered with or without repetition.

### Permutations without repetition

Take a look at the following example **Permutations** better to understand: you want to take a photo of your friends on your birthday. How many ways are there to line up your friends? For example, if you're a $ 6 person, there are $ 6 ways to pick the first person. For the second person there are still $ 5 options. For the third person there are still $ 4 $ possibilities, etc. In total you get $ 6 \ times 5 \ times 4 \ times 3 \ times 2 \ times $ possibilities. An abbreviation for this is the so-called **Faculty**. You're writing $ 6! = 6 \ times 5 \ times 4 \ times 3 \ times 2 \ times = 720 $. This is a permutation with no repetition.

### Permutations with repetition

In the example above, all persons can be distinguished. If $ k $ elements cannot be distinguished from the $ n $ elements of a set, then the number of possible arrangements is reduced. The formula is then $ N = \ frac {n!} {K!} $.

In the following we consider arrangements in which **a selection** is hit by $ k $ elements from a total of $ n $ elements.

### Variations

A variation is one **ordered sample**. So the order of the elements plays a role. Even with variations, we differentiate between variations with and without repetition.

### Variations without repetition

You have an urn in which there are $ n = 5 $ distinguishable balls. You draw $ (k = 3) $ three times without repetition. This means that the first draw gives you $ 5, the second $ 4, and the third $ 3. In total you get $ N = 5 \ times 4 \ times 3 = 60 $ possibilities.

In general, the formula $ N = \ frac {n!} {(N-k)!} $ Applies.

### Variations with repetition

We look again at the example with the urn and the $ n = 5 $ distinguishable balls. This time you draw with repetition. You draw the first ball and you have $ 5 $ opportunities. Then you repeat the drawing and you have $ 5 $ opportunities again. You also have $ 5 options on the third try. This gives a total of $ 5 ^ {3} = $ 125 possibilities.

In general, $ N = n ^ {k} $ applies.

As already said, the order of the dragging is taken into account in all of these examples. So you also differentiate between **Dragging with and without regard to the order**.

### Combinations

A combination is a **disordered sample**. The order of the objects will **not taken into account**. We again differentiate between combinations with and without repetition.

### Combinations without repetition

A nice example of this is the drawing of the lottery numbers on Saturdays. A selection of $ k = 6 $ from $ n = 49 $ balls is made.

Since the order in which the balls are drawn does not matter, these are combinations. The number of **Variations** can be determined quickly. There are $ \ frac {49!} {43!} $ Possible variations. These variations can now be divided into groups that contain the same numbers. For example, the drawings $ \ {1,23,35,44,27,15 \} $ and $ \ {22,1,35,44,27,15 \} $ are the same combination. In total, there are $ 6! $ Permutations for each selection of numbers. Therefore you get in total:

$ N = \ frac {49!} {43! \ Cdot 6!} = 13983816 $ or $ N = \ begin {pmatrix} 49 \ 6 \ end {pmatrix} $

In general you get:

$ N = \ begin {pmatrix} n \ k \ end {pmatrix} = \ frac {n!} {(N-k)! \ Cdot k!} $

The expression $ \ begin {pmatrix} n \ k \ end {pmatrix} $ is called **Binomial coefficient**. You read it as "$ n $ over $ k $".

### Combinations with repetition

With this selection you restore the initial situation after each "move". The formula here is:

$ N = \ begin {pmatrix} n + k-1 \ k \ end {pmatrix} $

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