Question

How do you evaluate \[{}^{8}{{P}_{3}}\] ?

Answer 1

Hint: Here \[P\] stands for Permutation, since we can calculate the value of permutation \[{}^{n}{{P}_{r}}\] as we know that formula for this in terms of \[n\] and \[r\] first we need to compare this to know the value of \[n\] and \[r\] after finding these values just put into the formula of permutation and that is \[\dfrac{n!}{(n-r)!}\].

Formula used:
 \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\]
Where, \[n!=n(n-1)(n-1)(n-3)........1\]

Complete step by step solution:
Since we have to calculate the value of \[{}^{8}{{P}_{3}}\],
Also, we know that \[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\]
Comparing this with the given question,
\[\Rightarrow n=8,r=3\]
Now putting these values in the formula
\[{}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}\]
\[\Rightarrow {}^{8}{{P}_{3}}=\dfrac{8!}{(8-3)!}\]
\[\Rightarrow \dfrac{8!}{5!}\]
Now we know that factorial is the product of all the numbers from \[1\] to that number itself
\[\Rightarrow \]This can be written as
\[\begin{align}
  & 8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \\
 & 5!=5\times 4\times 3\times 2\times 1 \\
\end{align}\]
Substituting these values in above term
\[\Rightarrow \dfrac{8!}{5!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}\]
On simplifying we get
\[\Rightarrow 8\times 7\times 6\]
\[\Rightarrow 336\]

Hence the value of \[{}^{8}{{P}_{3}}\] is \[336\].

Note: When we have to find the permutation or combination just recall the formula and compare it with the given question and substitute the value of the variables in the formula. Also in the calculation part \[8!\] can be written as \[8\times 7\times 6\times 5!\] then the \[5!\] will directly cancel out instead of writing all the stuff.