Variations without repetition

Where $A$ is a set of $n$ elements. The variations of $n$ elements taken from $k$ in $k$ are the number of arranged groups of the different elements of $A$, and it is represented as $V_{n,k}$.

For example,

Let's take the set $A=\{ a,b,c,d,e\}$. First of all, let's observe that, for example, the groups abc and bca are considered different, for they have the same letters ($a$, $b$ and $c$), but these are not in the same order. Then:

• The variations without repetition of these 5 elements taken in ones are: $a$, $b$, $c$, $d$ and $e$.
• The variations without repetition of these 5 elements taken in twos are: $ab$, $ba$, $ac$, $ca$, $ad$, $da$, $ae$, $ea$, $bc$, $cb$, $bd$, $db$, $be$, $eb$, etc...
• The variations without repetition of these 5 elements taken in threes are: $abc$, $acb$, $abd$, $adb$, $abe$, $aeb$, $bcd$, $cad$, etc...
• The variations without repetition of these 5 elements taken in fours are: $abcd$, $acbd$, $bacd$, $ebac$, $caed$, etc...
• The variations without repetition of these 5 elements taken in fives are: $abcde$, $abced$, $acbed$, $adbce$, $baced$, $bcdea$, etc...

As visible in the previous example, to count all the possibilities in each case could be very difficult, since there are many. But,it is possible to calculate it quickly by means of the following formula: $$\displaystyle V_{n,k}=\frac{n!}{(n-k)!}$$ where $n$ is the number of elements that set $A$ has, while $k$ is the number of elements that the group going to be constituted has.

Let's remember that $V_{n,k}$ is the quantity of variations without repetition of $n$ elements taken $k$ by $k$.

In the previous example,

Where $A=\{ a,b,c,d,e \}$ , then $n =5$.

Now, if we want to know the variations without repetition of these $5$ elements taken in threes, using the formula we see there are $60$: $$\displaystyle V_{5,3}=\frac{5!}{(5-3)!}=\frac{5 \cdot 4 \cdot 3 \cdot \not{2} \cdot 1}{\not{2} \cdot 1}=60$$