Answer
The required solution is $\frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=0$ .
Work Step by Step
The permutation and combination formulas respectively are:
${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
Applying these formulas into the provided expression, we get:
$\begin{align}
& \frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=\frac{7!}{3!\left( 7-3 \right)!}-\frac{7!}{3!\left( 7-3 \right)!} \\
& =0
\end{align}$
$\frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=0$
