# Combinations

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Every group of ${\displaystyle k}$ elements chosen from a set of ${\displaystyle n}$ elements in which ordering of the chosen elements is unimportant is called a combination of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements.

## Combinations without repetition

Ordering of elements does not play any role when the number of combinations is to be determined (i.e., groups ${\displaystyle ab}$ and ${\displaystyle ba}$ are equivalent combinations). That is why the number of combinations of the ${\displaystyle k}$th order is lower than the number of variations of the ${\displaystyle k}$th order from the same set of ${\displaystyle n}$ elements. The number of variations, which differ from each other just by ordering of their elements, is given by ${\displaystyle P(k)}$. Hence, the number of combinations of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements without repetition (it is denoted here by ${\displaystyle K(n;k)}$) is: ${\displaystyle K(n;k)={\frac {V(n;k)}{P(k)}}={\frac {n\,!}{k\,!\cdot (n-k)\,!}}=\left({\begin{array}{c}n\\k\end{array}}\right)}$ Examples with elements , and (${\displaystyle n=3}$)

• For ${\displaystyle k=1}$ is ${\displaystyle K(3;1)=3}$ and these three possibilities are:

• For ${\displaystyle k=2}$ is ${\displaystyle K(3;2)=V(3;2)/P(2)=6/2=3}$:

• For ${\displaystyle k=3}$ is ${\displaystyle K(3;3)=V(3;3)/P(3)=3/3=1}$,

so there is just one combination:

## Combinations with repetition

Combinations with repetition can include one element more times; hence, the maximal possible number of combinations of the ${\displaystyle k}$th order of ${\displaystyle n}$ elements with repetition (denoted by ${\displaystyle K^{W}(n;k)}$) is ${\displaystyle K^{W}(n;k)=\left({\begin{array}{c}n+k-1\\k\end{array}}\right)}$ Examples with elements , and (${\displaystyle n=3}$)

• For ${\displaystyle k=1}$ is ${\displaystyle K^{W}(3;1)=3}$ and the three possibilities are:

• For ${\displaystyle k=2}$ is ${\displaystyle K^{W}(3;2)=6}$:

Millions of Germans try every Saturday their luck in the lottery called Lotto. They choose 6 numbers from 49 and hope that, thanks to these 6 numbers, they will get rich. They base the choice often on various almost “mystical” numbers—numbers such as the date of somebody’s birthday, the birthday of their dog, numbers hinted by a horoscope and so on. How many possibilities for a choice of 6 numbers out of 49 actually exists? From 49 numbers (elements), exactly 6 is chosen. The order in which the numbers are chosen in unimportant—it does not matter whether one crosses first 4 and then 23 or vice versa. That means that ordering of elements is not taken into consideration. Therefore, permutations (simple reorderings of ${\displaystyle n}$ elements) and variations as well (ordering of elements matters) are not the right choice. The right concept is a combination. Nevertheless, there are still two possibilities—combinations with or without repetition. Since every number of the lottery ticket can be crossed just once, repetition of numbers (elements) is not possible and we use combinations without repetition. ${\displaystyle n=49\qquad k=6}$ ${\displaystyle K(n,k)=\left({\begin{array}{c}n\\k\end{array}}\right)={\frac {V(n,k)}{P(k)}}={\frac {n\,!}{k\,!\cdot (n-k)\,!}}}$ ${\displaystyle K(n,k)={\frac {49\,!}{6\,!\cdot (49-6)\,!}}=13\,983\,816}$ There is 13983816 possible combinations of 6 numbers from 49.