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Question: How do you evaluate \({}^{8}{{p}_{5}}\)?...

How do you evaluate 8p5{}^{8}{{p}_{5}}?

Explanation

Solution

Consider a set of elements A=x,y,z\text{A}=\\{x,y,z\\}. The arrangement of elements taken two at a time will be xy, xz, yx, yz, zx, and zy. These are the permutations of the elements of the set A, taken two at a time.
If n is the given number of elements and r is the number of elements taken at a time, the possible number of permutations nPr{}^{n}{{P}_{r}} is given by the equation
⇒nPr=n!(n−r)!\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} ……(1)

Complete Step by Step Solution:
The permutation is expressed using the notation npr{}^{n}{{p}_{r}}. By comparing this with 8P5, we have n=8n=8 and r=5r=5.
Before evaluating the value of 8P5, let us discuss how the equation (1) is derived. If the number of given elements is ‘n’, there will be n number of possibilities for a given place. Hence for n places, the number of permutations will be n×nn\times n.
But this includes repeated ones like xx, yy, etc. Therefore we have to take out the previous element at each place. This is equal to n×(n−1)×(n−2)...×1n\times \left( n-1 \right)\times \left( n-2 \right)...\times 1. And it is mathematically written as
⇒n!=n×(n−1)×(n−2)...×1\Rightarrow n!=n\times \left( n-1 \right)\times \left( n-2 \right)...\times 1
In case if the number of elements taken at a time is less than n, the number of permutations is now equal to
⇒nPr=n!(n−r)!\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
Where r is the number of elements taken at a time.
Therefore the permutation for the given condition will be
⇒8P5=8!(8−5)!\Rightarrow {}^{8}{{P}_{5}}=\dfrac{8!}{\left( 8-5 \right)!}
⇒8P5=8×7×6×5×4×3×2×13×2×1\Rightarrow {}^{8}{{P}_{5}}=\dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}

⇒8P5=6720\Rightarrow {}^{8}{{P}_{5}}=6720

Note:
We don’t really care about the order of the arrangement of elements in the case of permutations. For example, the arrangements xy and yx have the exact same elements but in a different order. They will give the same results whenever the order of arrangement is insignificant.
If we ignore the order of arrangement, we will have xy, xz, and yz. These are the combinations of the elements of the set A, taken two at a time. And the number of combinations is denoted as nCr^{n}{{C}_{r}}.
Therefore we should have a clear understanding of the differences between permutation and combination.