Let’s start with an exercise of permutations, the statement is as follows:
If you take four of the seven letters of the word CHANGOS to form a word (regardless of whether the word has or not sense), calculate the number of words that can be done with four letters with the conditions that we will see below.
For each of the paragraphs that we will see below, let’s assemble a box with four fields to visualize the exercise easier.
\begin{array}{| c | c | c | c |} \hline \qquad & \qquad & \qquad & \qquad \\ \hline \end{array}First example of permutations
The words must have the letter S.
What we do first is to place the letter that the paragraph already tells us in any field of the box, we will place it in the first field:
\begin{array}{| c | c | c | c |} \hline \quad S \quad & \qquad & \qquad & \qquad \\ \hline \end{array}And then what you do is fill in the fields with the other letters of the word CHANGOS, but how do you do that? First let’s observe that we already put the letter S in the field, and how many letters S do we have in the word CHANGOS? We have only one letter S. Therefore, we will write the number 1 in any of the fields, in this case as we put the S in the first field, we will put the 1 in the first field:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \qquad & \qquad & \qquad \\ \hline \end{array}And as the letter S is already occupied, that means that now we only have six letters left of the word CHANGOS, which means that in the next field will be any of the 6 letters, in the next field any of the 5 remaining letters and in the next field any of the 4 remaining letters:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \quad 6 \quad & \quad 5 \quad & \quad 4\quad \\ \hline \end{array}But oh surprise! It follows that the letter S we accommodated only in the first field, what we mean is that the letter S can occupy the same place in the second field, in the third field and in the fourth field, visually I refer to the following:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \quad 6 \quad & \quad 5 \quad & \quad 4\quad \\ \hline \end{array}
\begin{array}{| c | c | c | c |} \hline \quad 6 \quad & \quad 1 \quad & \quad 5 \quad & \quad 4\quad \\ \hline \end{array}\begin{array}{| c | c | c | c |} \hline \quad 6 \quad & \quad 5 \quad & \quad 1 \quad & \quad 4\quad \\ \hline \end{array}\begin{array}{| c | c | c | c |} \hline \quad 6 \quad & \quad 5 \quad & \quad 4 \quad & \quad 1\quad \\ \hline \end{array}So the number of ways in which we can order a four letter word that must have the letter S is calculated as follows:
1\cdot 6\cdot 5\cdot 4
But remembering that the S can take any place of the four fields, we have to multiply by 4:
1\cdot 6\cdot 5\cdot 4 \cdot 4 = 480
That means that there are 480 ways to write a four letter word if it must have the letter S.
Second example of permutations
The words have to start with G and end in a vowel.
What we do first is to place the letter G in the first of the four fields and then one of the two vowels was placed in the last field:
\begin{array}{| c | c | c | c |} \hline \quad G \quad & \qquad & \qquad & \quad A \quad \\ \hline \end{array}Then what was observed is that we only have one letter G and two vowels, therefore, in numbers it would look like this:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \qquad & \qquad & \quad 2 \quad \\ \hline \end{array}And because we already have two letters separated, that means that in any of the other two remaining fields will go any of the 5 remaining letters and in the next will go any of the 4 remaining letters:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \quad 5 \quad & \quad 4 \quad & \quad 2 \quad \\ \hline \end{array}Multiplying, we will obtain:
1 \cdot 5 \cdot 4 \cdot 2 = 40
The result is that 40 words can be done if you start with the letter G and end in a vowel.
Third example of permutations
The words have to start with the letter G and include the N.
First we will place each of the two letters so that they occupy one field each one, the G goes in the first field and the N can be placed in any of the three fields, let’s see:
\begin{array}{| c | c | c | c |} \hline \quad G \quad & \quad N \quad & \qquad & \qquad \\ \hline
\end{array}Now what was observed is that there are only 5 letters of the word CHANGOS, those 5 letters were ordered in the remaining boxes:
\begin{array}{| c | c | c | c |} \hline \quad 1 \quad & \quad 1 \quad & \quad 5 \quad & \quad 4 \quad \\ \hline \end{array}But it follows that the N can be in any of the three fields in which the G is not. Look, visually the N can be in any of the following three fields:
\begin{array}{| c | c | c | c |} \hline \qquad & \quad N \quad & \qquad & \qquad \\ \hline \end{array}\begin{array}{| c | c | c | c |} \hline \qquad & \qquad & \quad N \quad & \qquad \\ \hline \end{array}\begin{array}{| c | c | c | c |} \hline \qquad & \qquad & \qquad & \quad N \quad \\ \hline \end{array}So we have to multiply by 3:
1\cdot 1 \cdot 5 \cdot 4 \cdot 3 = 60
There are 60 words that can be done if you start with the letter G and include the N.
Fourth example of permutations
The word must contain the two vowels
It was visualized that because there are only two vowels, by logic, once one is used, the other vowel can no longer be used, therefore, the fields were accommodated for the two vowels in the following way:
\begin{array}{| c | c | c | c |} \hline \quad A \quad & \quad O \quad & \qquad & \qquad \\ \hline \end{array}\begin{array}{| c | c | c | c |} \hline \quad O \quad & \quad A \quad & \qquad & \qquad \\ \hline \end{array}Representing it with numbers, it is as follows:
\begin{array}{| c | c | c | c |} \hline \quad 2 \quad & \quad 1 \quad & \qquad & \qquad \\ \hline \end{array}Then the other five letters of the word CHANGOS were ordered in the other fields:
\begin{array}{| c | c | c | c |} \hline \quad 2 \quad & \quad 1 \quad & \quad 5 \quad & \quad 4 \quad \\ \hline \end{array}BUT HERE DOES NOT END THE EXERCISE! It follows that those two fields of the vowels can be placed in any order, so as they do not have an established position, we have to apply combinations to know the amount of (forgive the redundancy) combinations in which those two fields can be placed in the positions. We can take the calculator for (again the redundancy) calculate the number of combinations that can take two fields of the four possible or we can use the formula to calculate the combinations:
C(4,2) = 4\text{C}2 = \left ( \begin{array}{c} 4 \\ 2 \end{array} \right)= \cfrac{4!}{2!\left(4-2\right)!} = 6
But we have two vowels, it means that 6 must be multiplied by 2:
6 \cdot 2 = 12
The final multiplication will be as follows:
2\cdot 1 \cdot 5 \cdot 4 \cdot 12 = 480
That means that 480 words can be done if the word has to have the two vowels.
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