The **counting techniques** are a series of probability methods for counting the possible number of arrangements within a set or sets of objects. These are used when doing the accounts manually becomes complicated due to the large number of objects and/or variables.

For example, the solution to this problem is very simple: imagine that your boss asks you to count the last products that have arrived in the last hour. In this case, you could go and count the products one by one.

However, imagine that the problem is this: your boss asks you to count how many groups of 5 products of the same type can be formed with those that have arrived in the last hour. In this case, the calculation is complicated. For this type of situation, the so-called counting techniques are used.

These techniques are various, but the most important are divided into two basic principles, which are the multiplicative and the additive; permutations and combinations.

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**multiplicative principle**

**Applications**

The multiplicative principle, along with the additive principle, are basic to understand how counting techniques work. In the case of the multiplicative, it consists of the following:

Imagine an activity that involves a specific number of steps (we mark the total as «r»), where the first step can be done in N1 ways, the second step in N2, and the step «r» in Nr ways. In this case, the activity could be carried out with the number of shapes resulting from this operation: N1 x N2 x……….x Nr shapes

That is why this principle is called multiplicative, and implies that each and every one of the steps that are needed to carry out the activity must be carried out one after the other.

**Example**

Let’s imagine a person who wants to build a school. To do this, consider that the base of the building can be built in two different ways, cement or concrete. As for the walls, they can be made of adobe, cement or brick.

As for the roof, it can be made of cement or galvanized sheet. Finally, the final painting can only be done in one way. The question that arises is the following: How many ways do you have to build the school?

First of all, we consider the number of steps, which would be the base, the walls, the roof and the painting. In total, 4 steps, so r=4.

The following would be to enumerate the N:

N1= ways to build the base=2

N2= ways to build the walls=3

N3= ways to make the roof = 2

N4= ways of painting = 1

Therefore, the number of possible shapes would be calculated using the formula described above:

N1 x N2 x N3 x N4 = 2 x 3 x 2 x 1 = 12 ways to do school.

**additive principle**** **

**Applications**

This principle is very simple, and it consists in the fact that, in the case of several alternatives to carry out the same activity, the possible ways consist of the sum of the different possible ways of carrying out all the alternatives.

In other words, if we want to carry out an activity with three alternatives, where the first alternative can be done in M ways, the second in N ways and the last one in W ways, the activity can be done in: M + N +………+ w shapes.

**Example**

Let’s imagine this time a person who wants to buy a tennis racket. For this, he has three brands to choose from: Wilson, Babolat or Head.

When you go to the store you see that the Wilson racquet can be bought with two different handle sizes, L2 or L3 in four different models and can be strung or unstrung.

The Babolat racket, on the other hand, has three handles (L1, L2 and L3), there are two different models and it can also be strung or unstrung.

The Head racket, for its part, only comes with one handle, the L2, in two different models and only unstrung. The question is: How many ways does this person have to buy his racket?

M = Number of ways to select a Wilson racket

N = Number of ways to select a Babolat racket

W = Number of ways to select a Head racket

We carry out the multiplier principle:

M = 2 x 4 x 2 = 16 shapes

N = 3 x 2 x 2 = 12 shapes

W = 1 x 2 x 1 = 2 shapes

M + N + W = 16 + 12 + 2 = 30 ways to choose a racket.

To know when to use the multiplicative principle and the additive, you only have to look at whether the activity has a series of steps to be carried out, and if there are several alternatives, the additive.

**permutations**

**Applications**

To understand what a permutation is, it’s important to explain what a combination is so you can tell them apart and know when to use them.

A combination would be an arrangement of elements in which we are not interested in the position that each of them occupies.

A permutation, on the other hand, would be an arrangement of elements in which we are interested in the position that each of them occupies.

Let’s give an example to better understand the difference.

**Example**

Imagine a class with 35 students, and with the following situations:

The teacher wants three of his students to help him keep the class clean or deliver materials to the other students when needed.

The professor wants to name the class delegates (a president, an assistant, and a financier).

The solution would be the following:

Let’s imagine that Juan, María and Lucía are chosen by vote to clean the class or deliver the materials. Obviously, other groups of three people could have been formed, among the 35 possible students.

We must ask ourselves the following: is the order or position that each of the students occupies important when selecting them?

If we think about it, we see that it really is not important, since the group is going to take care of both tasks equally. In this case, it is a combination, since we are not interested in the position of the elements.

Now let’s imagine that Juan is elected as president, María as assistant and Lucía as financier.

In this case, would the order matter? The answer is yes, since if we change the elements, the result changes. That is to say, if instead of putting Juan as president, we put him as assistant, and María as president, the final result would change. In this case it is a permutation.

Once the difference is understood, we are going to obtain the formulas of permutations and combinations. However, first you have to define the term “n!” (jan factorial), since it will be used in the different formulas.

n!= to the product from 1 to n.

n!= 1 x 2 x 3 x 4 x………..x n

Using it with real numbers:

10!=1 x 2 x 3 x 4 x………x 10=3,628,800

5!= 1 x 2 x 3 x 4 x………x 5=120

The formula of the permutations would be the following:

nPr=n!/(nr)!

With it we will be able to find out the arrangements where the order is important, and where the n elements are different.

**combinations**

**Applications**

As we have commented before, the combinations are the arrangements where we do not care about the position of the elements.

Its formula is the following:

nCr=n!/(nr)!r!

**Example**

If there are 14 students who want to volunteer to clean the classroom, how many cleaning groups can be formed if each group has to be 5 people?

The solution, therefore, would be the following:

n=14, r=5

14C5 = 14! / (14 – 5 )!5! = 14! / 9!5! = 14 x 13 x 12 x 11 x 10 x 9!/ 9!5!= 2002 groups

**solved exercises**

**Exercise 1**

Natalia is asked by her mother to go to a grocery store and buy her a soft drink to cool off. When Natalia asks her clerk for the drink, he tells her that there are four flavors of soft drinks, three types and three sizes.

The flavors of soft drinks can be: cola, lemon, orange and mint.

The types of cola can be: normal, without sugar, without caffeine.

The sizes can be: small, medium and large.

Natalia’s mother did not specify what type of soft drink she wanted. How many ways does Natalia have to buy the drink?

**Solution**

M = Number of size and type that you can select when choosing the cola drink.

N = Number of size and type that you can select when choosing the lemon soda.

W = Number of size and type that you can select when choosing the orange soda.

Y = Number of size and type that you can select when choosing the mint drink.

We carry out the multiplier principle:

M = 3×3 = 9 shapes

N = 3×3 = 9 shapes

W = 3×3 = 9 shapes

Y = 3×3 = 9 shapes

M + N + W + Y= 9 + 9 + 9 + 9 = 36 ways to select the soda.

**Exercise 2**

A sports club announces free access workshops for children to learn to skate. 20 children enroll, so they decide to divide them into two groups of ten people so that the instructors can give the classes more comfortably.

In turn, they decide to draw a lot in which group each child will fall into. How many different groups could a child enter?

**Solution**

In this case, the way to find an answer is through the combination technique, whose formula was: nCr=n!/(nr)!r!

n = 20 (number of children)

r = 10 (group size)

20C10 = 20! / (20 – 10)!10! = 20! / 10!10! = 20 x 19 x 18 x 17 x 16 x 15x 14x 13x 12x 11x 10!/ 10!10!= 184,756 groups.

**References**** **

Jeffrey, R.C. *Probability and the Art of Judgment,* Cambridge University Press. (1992).

william feller, *“An Introduction to Probability Theory and Its Applications*«, (Vol 1), 3rd Ed, (1968), Wiley

Finetti, Bruno de (1970). *“Logical foundations and measurement of subjective probability”*. Psychological Act.

Hogg, Robert V.; Craig, Allen; McKean, Joseph W. (2004). *Introduction to Mathematical Statistics* (6th ed.). Upper Saddle River: Pearson.

Franklin, J. (2001) *The Science of Conjecture: Evidence and Probability Before Pascal,*Johns Hopkins University Press.