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Question

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

Find the value of 8P8{^8}{P}{{_8}}

Explanation

Solution

In this question, we need to find the value of 8P8{^8}{P}{{_8}} . Permutations is the choice of arranging rr things from the set of nn things that is also without replacement. By using the formula of the permutation we can find the value of 8P8{^8}{P}{{_8}}
Formula used :
nPr=(n!)(nr)!{^n}{P}{{_r}}= \dfrac{\left( n! \right)}{\left( n – r \right)!}
Where,
P =PermutationP\ = \text{Permutation}
n=number of objectsn = \text{number of objects}
r =Number of objects selectedr\ = \text{Number of objects selected}
W. K. T :
0! =10!\ = 1
n!=n(n1)(n2)(n3)n! = n\left( n – 1 \right)\left( n – 2 \right)\left( n – 3 \right)\ldots

Complete step-by-step solution:
8P8{^8}{P}{{_8}},
Here n=8n = 8 and r=8r = 8
By expanding ,
We get,
8P8=(8!)(88)!{^8}{P}{{_8}}= \dfrac{\left( 8! \right)}{\left( 8 – 8 \right)!}
By simplifying,
We get,
8P8=8!0!{^8}{P}{{_8}}= \dfrac{8!}{0!}
We also know that 0!0! Is 11
By substituting,
We get,
8P8=8!1{^8}{P}{{_8}}= \dfrac{8!}{1}
By expanding,
We get,
W. K. T :
n!=n(n1)(n2)(n3) n! = n\left( n – 1 \right)\left( n – 2 \right)\left( n – 3 \right)\ldots\
Thus by expanding ,
We get,
8P8=8×7×6×5×4×3×2×1{^8}{P}{{_8}}= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
By multiplying,
We get,
8P8=40320{^8}{P}{{_8}}= 40320
Thus the value of 8P8=40320{^8}{P}{{_8}}= 40320
Final answer :
The value of 8P8=40320{^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.