# How many combinations on a 17x17 cube

## Calculate the number of possibilities / events (combinatorics)

In the calculation of probability, you will surely have to work out at some point how many possibilities or arrangements there are in an experiment. In concrete terms: How many possible events are there? In order to calculate this, it always depends on how the experiment is set up:

**Overview**

- Arrangements
- Number of possible events in an arrangement
- E.g. sit 5 people on 5 chairs

- Place 10 cars in 10 parking spaces

- Number of possible events in an arrangement with the same objects
- E.g. 3 VW's and 2 Volvos in 5 parking spaces

- Order of drawing 4 red and 2 blue balls

- Number of possible events in an arrangement
- Choose
- Considering the order
- Number of possible events without "replacement" or multiple selection
- E.g .: draw 3 out of 5 balls, whereby it is important which one is drawn first and which one last

- Number of possible events with "replacement" or multiple selection
- E.g. combination lock with 3 setting positions (3 rings on which you turn the number) and 10 numbers each.
- A binary number has 2 states (0 and 1).

- Number of possible events without "replacement" or multiple selection
- Without considering the order
- Number of possible events without "replacement" or multiple selection
- e.g. Lotto 6 out of 49, so you draw 6 balls out of 49
- Multiple toss of a coin, whereby the number of possibilities should be calculated, for example if two heads should occur.

- Number of possible events with "replacement" or multiple selection
- E.g. 4 balls are drawn from a pot of 6 balls, the ball is put back immediately after each time. How many options are there?

- Number of possible events without "replacement" or multiple selection

- Considering the order

### Number of possible events in an arrangement

This is the case if, for example, you have 5 people and want to calculate how many options there are to place them next to each other. This is relatively easy to calculate, you just take the factorial of the number of people or objects that you want to arrange. It is important that all objects are distinguishable. n is the number of objects:

n!

**Examples of use:**

- Seat 5 people on 5 chairs
- Place 10 cars in 10 parking spaces

**Exercise to practice:**

You want to know how many ways there are to seat your 10 birthday guests on the chairs at the table. | Fade in | |

Solution: To calculate this, take the factorial of 10, i.e. 10! = 3628800. So there are 3.6 million possibilities !!!! |

### Number of possible events in an arrangement with the same objects

So if you have several objects, some of which are the same and you want to know how to arrange them, you can calculate it as follows:

- Take the total faculty of the objects, i.e. how many there are
- Divide this by the factorial of all the same objects, so you have, for example, 6 balls, 4 of which are equal and two more equal, then you divide by 4! · 2 !.

**example**: Do you have **n balls and draws** one after the other but of these are k_{1} red, k_{2} black, k_{3} blue ... so they are the same. Then you calculate it like this:

**Examples of use:**

- 3 VWs and 2 Volvos in 5 parking spaces (n = 5, k1 = 3, k2 = 2)
- Order of drawing 4 red and 2 blue balls (n = 6, k1 = 4, k2 = 2)

**Exercise to practice:**

You want to design a new flag with stripes, for this you want to make 6 stripes, 3 of which should be red and 3 white. How many options are there? | Fade in | |

Solution: To calculate this, take the factorial of 6, i.e. all strips, and divide it by the faculties of the same strips, i.e. 3! and 3 !. It looks like this: 6! :( 3! 3!) = 20. So there are 20 options. |

### 2.1 Considering the order

By considering the sequence, one understands that it is also important which event occurred when.

### Number of events without "covering" or multiple selection with sequence

If you are to calculate the number of possible events, whereby one cannot "put back" an event twice, you can always think of it as an arrangement problem, so how many possibilities are there to arrange these combinations, then you do it like this:

- Again, take the faculty of the total number of objects that are available for selection
- You then divide that by the factorial of the number of objects that are left over, i.e. not selected. For example, if you choose 3 out of 5 balls, you divide by 2 !, since there are 2 balls left.

So in general (n is the total number of balls and k is the number of selected balls):

**Examples of use:**

- Draw 3 of 5 balls, it is important which one is drawn first and which one last. So it makes a difference whether, for example, a blue ball was drawn first and then the red ball or vice versa, that is, considering the sequence, 2 different results.

**Exercise to practice:**

You draw 2 balls from an urn with 4 balls, which are all different colors, without putting them back. It is important which ball was drawn first and which second, that makes a difference for you (e.g. if you draw red and then blue, the result is different for you than if you draw blue and then red) | Fade in | |

Solution: So you divide the factorial of 4 by the factorial of the remaining spheres, i.e. 2 !. The result is then: 4!: 2! = 12. So there are 12 options. |

### Number of events with "move back" or multiple selection with sequence

If you should calculate the number of possibilities, if you have to choose which objects from objects, but can also choose objects several times (e.g. after each pulling the ball back into the lottery pot), whereby the order is also important, then you do this by simply take the number of total objects up to the number you choose.

(n is the number of elements (or possibilities) and k is the number of "draws")

n^{k}

**Examples of use:**

- Combination lock with 3 setting positions (3 rings on which you turn the number) and 10 numbers each. (n = 10 and k = 3). You can, for example, set the 9 again at any point of the lock, therefore with multiple selection.
- A binary number has 2 states (0 and 1). With a sequence of 10 numbers, 2 to the power of 10 different variations can arise. (n = 2 and n = 10)

**Exercise to practice:**

You want to crack the password of a cell phone, which has 4 digits and only consists of numbers, so there are 10 options per digit of the password (0, 1, 2, 3 ... 9). How many combinations are there? | Fade in | |

Solution: So you take the number of possibilities per position multiplied by the number of positions, i.e. 10 |

### 2.2 Without considering the order

Without considering the order, it does not matter whether one ball was drawn first and then the other or vice versa. Both events are synonymous. The following calculations are without consideration of the order:

### Number of events without "covering" or multiple selection without sequence

(on the subject of binomial coefficients, click HERE) Should you calculate the number of possible events, whereby one does not "put back", i.e. an event must not occur twice (so you calculate how many possible combinations there are) without considering the order, do you do it like this (n is the number of elements and k is the number of selections):

**Application example:**

- Lotto 6 out of 49, so you draw 6 balls from 49, the order does not matter whether the 3 is drawn first or last, it makes no difference. (n = 49 and k = 6)
- Multiple toss of a coin, whereby the number of possibilities should be calculated, for example if two heads should occur. (n = number of throws and k = number of head throws)

**Exercise to practice:**

You are playing the lottery and want to know how many options there are to choose 6 out of 49 numbers. | Fade in | |

Solution: You share the faculty of 49! by the factorial of (49-6)! times 6 !. The result is then: 49! :( 43! · 6!) = 13983816. That is a rounded 14 million possibilities! |

### Number of events with "move back" or multiple selection without sequence

The number of possible events, whereby "backs up" or the results may occur multiple times, regardless of the sequence. The calculation looks like this (n is the total number of balls and k is the number of balls you choose):

**Application example:**

- 4 balls are drawn from a pot of 6 balls, the ball is put back immediately after each time.

**Exercise to practice:**

You draw 3 balls from an urn with 6 different balls. Each drawn ball is put back directly. The order in which the balls are drawn does not matter (e.g. first blue then red is the same as first red then blue). | Fade in | |

Solution: The result is then: (6 + 3-1)!: ((6-1)! 3!) = 56. So there are 56 ways to draw these 3 balls out of 6. |

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