Question

How do you evaluate ${}_{7}{{P}_{7}}$?

Answer 1

We can infer by looking at the question that this is from Permutations and Combinations chapter. But for now, let us only look at permutation. Permutation is defined as the different arrangements which can be made by taking some or all of a number of things. Mathematically, it is the number of ways of arranging $n$ distinct objects in a row taking $r\left( 0 < r < n \right)$ at a time is denoted by $P\left( n,r \right)$ or ${}^{n}{{P}_{r}}$.
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .

Complete step by step answer:
${}_{n}{{P}_{r}}$ is the same as ${}^{n}{{P}_{r}}$.
The permutation which is given to us is ${}_{7}{{P}_{7}}$. It is the same as ${}^{7}{{P}_{7}}$.
So now let us evaluate it.
Upon comparing the given permutation with the standard form, we can conclude the following:
$\begin{aligned} & n=7 \\\ & r=7 \\\ \end{aligned}$
$r$should not be greater than $n$. It can be less than or equal to $n$.
$r,n$ are not mentioned in the question. We have taken up so as to simplify the question and arrive at the answer easily.
$\Rightarrow {}_{7}{{P}_{7}}={}^{7}{{P}_{7}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{7!}{\left( 7-7 \right)!}=\dfrac{7!}{0!}$ .
Here we must know that $0!$is equal to $1$ but not $0$.
So upon solving, we get the following :
$\Rightarrow {}_{7}{{P}_{7}}={}^{7}{{P}_{7}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{7!}{\left( 7-7 \right)!}=\dfrac{7!}{0!}=7!$.
We know that $7!=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$.
$\Rightarrow {}_{7}{{P}_{7}}={}^{7}{{P}_{7}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{7!}{\left( 7-7 \right)!}=\dfrac{7!}{0!}=7!=5040$

$\therefore$Hence, upon evaluating ${}_{7}{{P}_{7}}$, we get $5040$.

Note: It is very important to learn all the formulae and definitions present in this chapter. We should be very clear in each concept and not get confused between permutation and combination. Both of them seem similar but are yet so different.