Find the value of \[{^8}{P}{{_8}}\]

Hint: In this question, we need to find the value of \[{^8}{P}{{_8}}\] . Permutations is the choice of arranging \[r\] things from the set of \[n\] things that is also without replacement. By using the formula of the permutation we can find the value of \[{^8}{P}{{_8}}\]
Formula used :
\[{^n}{P}{{_r}}= \dfrac{\left( n! \right)}{\left( n – r \right)!}\]
Where,
\[P\ = \text{Permutation}\]
\[n = \text{number of objects}\]
\[r\ = \text{Number of objects selected}\]
W. K. T :
\[0!\ = 1\]
\[n! = n\left( n – 1 \right)\left( n – 2 \right)\left( n – 3 \right)\ldots\]

Complete step-by-step solution:
\[{^8}{P}{{_8}}\],
Here \[n = 8\] and \[r = 8\]
By expanding ,
We get,
 \[{^8}{P}{{_8}}= \dfrac{\left( 8! \right)}{\left( 8 – 8 \right)!}\]
By simplifying,
We get,
 \[{^8}{P}{{_8}}= \dfrac{8!}{0!}\]
We also know that \[0!\] Is \[1\]
By substituting,
We get,
 \[{^8}{P}{{_8}}= \dfrac{8!}{1}\]
By expanding,
We get,
W. K. T :
\[n! = n\left( n – 1 \right)\left( n – 2 \right)\left( n – 3 \right)\ldots\ \]
Thus by expanding ,
We get,
\[{^8}{P}{{_8}}= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
By multiplying,
We get,
\[{^8}{P}{{_8}}= 40320\]
Thus the value of \[{^8}{P}{{_8}}= 40320\]
Final answer :
The value of \[{^8}{P}{{_8}}= 40320\]


Note: We should not confuse combinations and permutations. Permutations are defined as the number of ways of arranging items in a particular order. In other words, selection and arrangement of subsets is known as permutation whereas the non fraction order of selection is known as combination. Combinations is a way of selecting items from a collection whereas the order of selection does not matter. The rule of the permutation is anything permute itself is equivalent to itself factorial. In combination, the order of the object is not important.