Question

What is the value of \[\dfrac{{{}^7{P_6}}}{{{}^7{P_4}}}\] ?

Answer 1

Hint: First, we will learn about what is permutation. In Statistics, a permutation is nothing but the number of possible arrangements in a set of objects with regard to the order of the arrangement.
Let us understand it easily with an example. We consider a set of data containing three letters \[a,b,{\text{ }}\] and $c$. And our task is to arrange 2 letters in all the possible ways. Hence, our complete possible list will be $ab$, $ac$, $ba$, $bc$, $ca$, $cb$.
Formula to be used:
The formula to calculate the permutation is as follows.
\[{}^n{P_r} = n(n - 1)(n - 2).......(n - r + 1)\]
$ = \dfrac{{n!}}{{(n - r)!}}$ (! Is a mathematical symbol called factorial )
Where, $n$ denotes the number of objects from which the permutation is formed and $r$ denotes the number of objects used to form the permutation.

Complete step-by-step solution:
We shall first use the above formula for our example set containing three letters $a,b$ and$c$ .
Here, $n$ is$3$ since the permutation was formed from 3 letters ($a,b$ and $c$).
Also, $r$ is $2$ since we use $2$ letters to form the required permutation.
By using the above formula, \[{}^3{P_2} = \dfrac{{3!}}{{(3 - 2)!}}\]
$ = 3!$
$ = 3 \times 2 \times 1 = 6$
Hence 6 possible ways are there the example.
Next, getting into the question, we need to calculate \[\dfrac{{{}^7{P_6}}}{{{}^7{P_4}}}\]
We shall find these two permutations separately.
1)${}^7{P_6}$ :
Here $n = 7$ and $r = 6$
By using above formula, ${}^7{P_6} = \dfrac{{7!}}{{(7 - 6)!}}$ $ = 7!$ … (1)
(2) \[{}^7{P_4}\]:
Here n=7 and r=4.
By using above formula, ${}^7{P_4} = \dfrac{{7!}}{{(7 - 4)!}}$ $ = \dfrac{{7!}}{{3!}}$…(2)
Hence, by using equations (1) and (2),
\[\dfrac{{{}^7{P_6}}}{{{}^7{P_4}}}\]=$\dfrac{{7!}}{{\dfrac{{7!}}{{3!}}}}$
$ = 7! \times \dfrac{{3!}}{{7!}}$
$ = 3!$ (by cancelling $7!$ in both numerator and denominator)
$ = 3 \times 2 \times 1$ (by expanding $3!$ )
$ = 6$
Therefore, \[\dfrac{{{}^7{P_6}}}{{{}^7{P_4}}} = 6\].


Note: The term permutation refers to arrange the objects whereas the term combination refers to the process of selecting the objects. Both operations are different to each other. permutation is represented by ${}^n{P_r}$ and combination is represented by ${}^n{P_r}$.