How do you find the value of the permutation $P\left( {12,0} \right)$?

Hint: As we have to find the value of the permutation, we have to use the formula of the permutation i.e., $P\left( {n,r} \right) = {}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. Here, we are given the value of $n$ and $r$. Substitute the values in the formula and simplify to get the desired result.

Complete step by step answer:
The given permutation is $P\left( {12,0} \right)$.
A permutation of a set is, loosely speaking, an assembly of its members into a sequence or linear order, or a rearrangement of its elements if the set is already arranged. The term "permutations" often refers to an ordered set's act or method of modifying the linear order.
The formula of the permutation is,
$P\left( {n,r} \right) = {}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
We have to find the value of the permutation.
Comparing the given permutation with the formula, we get $n = 12$ and $r = 0$,
Substitute these values in the formula,
$ \Rightarrow P\left( {12,0} \right) = \dfrac{{12!}}{{\left( {12 - 0} \right)!}}$
Simplify the terms,
$ \Rightarrow P\left( {12,0} \right) = \dfrac{{12!}}{{12!}}$
Cancel out the common factor from numerator and denominator,
$ \Rightarrow P\left( {12,0} \right) = 1$

Hence, the value of the permutation $P\left( {12,0} \right)$ is 1.

Note: Here when solving these problems, if there is some term with $n$ letters and a letter repeats for $r$ times in it, then it can be organized in $\dfrac{{n!}}{{r!}}$ number of ways. If a set of letters are replicated for a different number of times, such as $n$ letter word and ${r_1}$ repeated items, ${r_2}$ repeated items,...... ${r_k}$ repeated items, then it is arranged in $\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_k}!}}$ number of ways.