Combinatorics is the branch of mathematics that deals with the number of possible ways to combine different elements.

Combinatorics can be used as a tool in probability theory.
The combinations found in combinatorics can be used as outcomes in probability calculations.

$P\left(Event\right)=\frac{favourablecombinations}{possiblecombinations}$P(Event)=(favourable combinations)/(possible combinations)

Concepts

“Either–or” (Addition principle)

If something is “either–or”, the numbers should be added together.

Example: There are two bowls with balls. In one bowl there are 2 balls (one black and one white). In the other bowl there are 3 balls (one green, one blue, and one red). How many possible choices are there if you take a ball either from bowl 1 or from bowl 2?

Solution by calculation:

There are 2 + 3 possibilities = 5 possibilities

“Both–and” (Multiplication principle)

If something is “both–and”, the numbers should be multiplied.

Example: There are two bowls with balls. In one bowl there are 2 balls (one black and one white). In the other bowl there are 3 balls (one green, one blue, and one red). How many possible combinations are there if you take a ball both from bowl 1 and from bowl 2?

Solution by calculation:
There are 2·3 possibilities = 6 possibilities

Solution using a tree diagram:

Som both the calculation and the tree diagram show, there are 6 possible combinations.

Formulas

Addition: $a+b=number of combinations$ (Either–or)
Multiplication: $a·b=number of combinations (Both and)$

n = Number of elements to choose from.
r = Number of elements chosen

Ordered with replacement: $n^r=number of combinations$number of combinations=n^r

Ordered without replacement:$P=\frac{n!}{(n-r)!}$number of combinations=n!/((n-r)!)

Unordered with replacement: $A(n,r)=\frac{(n+r-1)!}{r!·(n-1)!}$number of combinations=((n+r-1)!)/(r!*(n-1)!)

Unordered without replacement: $K(n,r)=\frac{n!}{r!\cdot (n-r)!}$number of combinations=(n!)/(r!*(n-r)!)