# Combinatorial Mathematics?

Posted July 4, 2022

on:Ivan. Jun 26, 2022

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# 1. Permutación, Disposition, Combination

In this article we are going to approach: Simple permutation, Permutation with repetition, Simple disposition, Disposition with repetition, Simple combination, Binomial Coefficients,

Considering:n = elementsk = available positions,

**Simple Permutation**

We had this permutation when there are n possible spaces and n elements. For example, if I had 7 positions on the podium and 7 horses, every horse will arrive in one of these positions.

How can I calculate all the possible combinations?

I can take 7(horses) and calculate the **factorial of 7**.

That’s mean = 7 * 6 * 5 * 4 * 3 * 2 * 1. The factorial is indicated with !.

So 7. Why?

Every horse can arrive at every position: The first can be occupied by all the seven horses. The second can be occupied by six horses because one is already in the first position. The third can be occupied by 5 horses because there’s one horse in the first position and one in the second.…

Those are the simple permutation! Pretty easy! For now…

**Permutation with repetition**

But if I have repeated elements how can I do it?

For example, I have 3 marble, two yellows and one red. The two-position occupied by the two yellow marbles are the same. How can I calculate in this case?

Doing the factorial of 3, for the 3 marbles, divided by the factorial of every group of marbles, in this case, 2!1! two yellows and one red.

Factorial 3 divided by factorial 2 * factorial1

**Simple Disposition**

When I want to order more elements than the available position. For example, if I want to order 3 elements in 7 positions. We do the same things as Simple Permutation but only for the requested positions,

In this case: 7 x 6 x 5.

**Disposition with Repetition**

When the results can be repeated every time, for example, when we throw a die.

Every time the possible number is the same, from 1 to 6.

Considering 5 throws the possible results are 6⁵.n = 6 (possible elements)k = 5 (number of lunch)

**Simple Combination**

In this case, we had to consider a group of elements n. Every element of the group is different. So we wanna say all the possible combinations in this group, composed of 8 elements, that contain only 3 elements.

Considering that every element is unique, a group can’t exist with the same element, and two groups with the same element but in different positions are the same group.

For example, considering: {a,b,c,d,e,f,g,h} so n = 8

I have to calculate every subset composed of 3 elements.{a,b,c} and {c,a,b} are the same set.

How can I proceed? Factorial 8 divided by factorial 3 * factorial 5

**Binomial Coefficients**

Count the subset of k beginning with n elements, basically it count the simple combination, it’s the same formula we have used before.

**Binomial Coefficient formula**

Don’t you remember what n and k are? Go back to the beginning of the articles!

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