What are some examples of permutations

Combinatorics: formulas, examples, tasks

Math> Probability and Statistics

In math in the Combinatorics, a sub-area of ​​stochastics, it is about determining the number of possible Arrangements or Choose of Objects.

Depending on whether you want to calculate different arrangements, options or both, there are different arithmetic operations. The following decision tree helps to decide which calculation is required for a specific task:

Decision tree combinatorics

In the following, we will go through the different calculation options and show you the variants of the combinatorics using various examples and tasks.

1st possibility: No selection is made

First of all, we have to ask ourselves whether the task is talking about a selection of objects or the total number of objects is meant (= no selection).

Becomes no choice we use the to calculate the various possible arrangements for the objects permutation. A distinction is made between a number of objects that are all distinguishable (= permutation without repetition) and a number of objects that are partially indistinguishable from one another (= permutation with repetition).

Permutation without repetition

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In order to calculate the number of different possible combinations of $ n $ distinguishable objects, one calculates:

$ \ Large {n!} $

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There are six different colored balls in an urn. How many ways are there to arrange the balls in a row?

$ n ~ = ~ 6 $

$ n! ~ = ~ 1 \ times 2 \ times 3 \ times 4 \ times 5 \ times 6 ~ = ~ 720 $

There are a total of 720 possibilities.

Permutation with repetition

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The number of possible combinations of $ n $ objects, of which $ k $ objects are identical, is calculated as follows:

$ \ Large {\ frac {n!} {K!}} $

If several objects are identical, the following applies:

$ \ Large {\ frac {n!} {K_1! \ cdot k_2! ...}} $

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There are three green and two yellow balls in an urn. How many ways are there to arrange the balls in a row?

$ \ Large {\ frac {n!} {K!} ~ = ~ \ Frac {5!} {3! \ cdot 2!} ~ = ~ \ frac {1 \ cdot 2 \ cdot 3 \ cdot 4 \ cdot 5} {(1 \ cdot 2 \ cdot 3) \ cdot (1 \ cdot 2)} ~ = ~ \ frac { 120} {12} ~ = ~ 10} $

There are $ 10 options.

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2nd possibility: A selection is made

Will be a selection hit by objects from a total set, we calculate the combination or the variation. The permutation doesn't help us in this case.

The combination indicates the number of possibilities to select a certain amount of objects from a larger total amount.

The variation indicates how many possibilities there are of arranging a certain selection of objects. The variation therefore takes two things into account: On the one hand, there are different options for making a selection. On the other hand, this selection can be arranged differently.

Combination without repetition

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In order to calculate how many possibilities there are to select $ k $ objects from a total of $ n $ objects, one calculates:

$ \ Large {\ binom {n} {k}} $

Spoken: "n over k" or "k from n "

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In the lottery, six numbers are chosen from a total of $ 49. How many options are there? You can calculate the combinations like this:

Number of selected objects $ k ~ = ~ 6 $

Number of total amount of objects $ n ~ = ~ 49 $

Calculation of the combination: $ \ Large {\ binom {n} {k} ~ = ~ \ binom {49} {6}} ~ = ~ $ 13,983,816

There are 13,983,816 (nearly 14 million) choices.

Combination with repetition

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In order to calculate how many possibilities there are to select $ k $ objects from a total of $ n $ objects, whereby the objects can be selected several times, one calculates:

$ \ Large {\ binom {n + k - 1} {k}} $

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There are six different colored balls in a vessel. Three of the balls are drawn, the drawn ball being replaced after each move (= with repetition).

Number of selected objects $ k ~ = ~ 3 $

Number of total amount of objects $ n ~ = ~ 6 $

Calculation of the combination: $ \ Large {\ binom {n + k - 1} {k} ~ = ~ \ binom {6 + 3 - 1} {3} ~ = ~ \ binom {8} {3}} ~ = ~ $ 56

There are 56 options.

Variation without repetition

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In order to calculate the number of possible combinations of a selection of $ k $ objects from a total number of $ n $ objects, we use the following formula:

$ \ Large {\ frac {n!} {(N - k)!}} $

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There are six different colored balls in a box, four of which are drawn. How many ways are there to order the selection of four balls?

$ \ Large {\ frac {n!} {(N - k)!} = \ Frac {6!} {(6 - 4)!} = \ Frac {6!} {2!} \ Frac {1 \ cdot 2 \ times 3 \ times 4 \ times 5 \ times 6} {1 \ dot 2} = \ frac {720} {2} = 360} $

So there are a total of $ 360 $ opportunities to draw four balls from a set of six balls and arrange them in the most varied of combinations.

Variation with repetition

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About the variation with repetition To calculate a selection of $ k $ objects from a total of $ n $ objects, this formula is required:

$ \ Large {n ^ k} $

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There are six different colored balls in a box, four of which are drawn. After each draw, the drawn ball is put back into the urn. How many possible combinations of drawn balls are there? Calculate the combinations.

Number $ n $ of all objects: $ 6 $

Number $ k $ of the selected objects: $ 4 $

$ \ Large {n ^ k = 6 ^ 4 = 1296} $

So there is a total of $ 1296 ways to pull four balls from a set of six balls with replacement and to arrange them in various combinations.

Now you know all the formulas in combinatorics and can calculate the permutation, combination and variation. Test your newly learnedKnowledge on the subject of combinatorics with ourExercises to combinatorics!

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There are ten balls in an urn, three of which are red and three are yellow. The other balls are of different colors. How many ways are there to arrange these balls in a row?

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