**Combinatorial without repetition is understood as the different sets that can be formed with «n» elements, selected from x to x. Each set must be differentiated from the previous one in at least one of its elements (the order does not matter) and these cannot be repeated.**

The combinatorial without repetition is commonly used in **statistics** and mathematics. This fits many real-life situations and its application is quite simple.

Consider, for example, a student who has a 4-question exam. Of the 4 questions you have to choose three, how many different combinations could the student make? If we reason a little we would see (without applying the formula) that the student could choose how to answer the 3 questions in four different ways.

- Set / option 1: Answer questions 1,2,3.
- Set / option 2: Answer questions 1,2,4.
- Set / option 3: Answer questions 1,3,4.
- Set / option 4: Answer the questions 2,3,4.

As we see, the student can form 4 sets (n) of 3 elements (x). Therefore, the combinatorial without repetition tells us how to form or group a finite amount of data / observations, in groups of a certain amount without any of the elements can be repeated in each group. This is the main difference between the **combinatorial with repetition** (the elements in each group can be repeated) and the combinatorial without repetition (no element can be repeated in each group)

To highlight in this example, which is a case of combinatorics without repetition, since the student cannot choose to ask any of the questions more than once. Therefore the elements of the sets cannot be repeated.

In the previous case, since the total number of elements is small and the quantity of the set is high, the quantity of options is small and can be deduced in a simple way without applying the formula. In the case of applying the formula directly, the numerator would be 24 (4 * 3 * 2 * 1) and the denominator would be 6 (3 * 2 * 1 * 1) with which we would arrive at the calculation in the same way without thinking about how we could group those four questions into sets of three.

## How to calculate the combinatorial without repetition?

The combinatorial formula without repetition is:

n = Total observations

x = Number of selected elements

**Example of ** combinatorics without repetition

Imagine a military squad of 12 soldiers. The army captain wants to form groups of 2 soldiers to infiltrate behind enemy lines through different points, how many different groups could he form?

To solve the problem, we must first identify the total number of elements. In this case there are 12 soldiers in total, so we already have our n. As the captain wants groups of 2, we already know what our x is. Knowing this, we could substitute in the formula and have the number of combinations of groups of 2.

n = 12

x = 2

Applying the factorial to the denominator we would have 12 * 11 * 10 * … * 1 = 479.001.600. For the denominator we have 2 * 1 * 10 * 9 * 8… * 1 = 7,257,600. Our combinatorial number is = 479.001.600 / 7.257.600 = 66.

As we can see, the captain can form 66 different pairs of soldiers among the 12 he has.