The value of ${ }^7 P_3-3\left({ }^6 P_2\right)$ is equal to

Question

Aicharya · Accepted Answer

We have, $ \begin{array}{ll} & { }^7 P_3-3{ }^6 P_2 \ & \frac{7 !}{4 !}-3 \cdot \frac{6 !}{4 !} \ \Rightarrow \quad & \frac{7 !}{4 !}-\frac{3 \cdot 6 !}{4 !} \ & =\frac{6 !}{4 !}(7-3)=\frac{6 !}{4 !} 	imes 4 \ & =\frac{6 ! 	imes 4}{4 	imes 3 !}=\frac{6 !}{3 !}={ }^6 P_3 \end{array} $

The value of ${ }^7 P_3-3\left({ }^6 P_2\right)$ is equal to

Solution