Answer
Option (c) is correct
The value of \(^7P_3=210\)
Step-by-step explanation:
Given : \(^7P_3\)
We have to find the value of \(^7P_3\)
Consider \(^7P_3\)
Permutation is defined as number of possibilities for choosing an ordered set of r objects from a total of n objects.
\(^nP_r=\frac{n!}{\left(n-r\right)!}\)
put n = 7 and r = 3
We have,
\(^7P_3=\frac{7!}{\left(7-3\right)!}\)
Simplify, we have
\(=\frac{7!}{4!}\)
\(\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:n>m\)
\(\frac{7!}{4!}=7\cdot \:6\cdot \:5=210\)
Thus, The value of \(^7P_3=210\)
