Answer
Yes, the statement is true.
Work Step by Step
$\displaystyle \left(\begin{array}{l}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}$
$\displaystyle \left(\begin{array}{l}n\\n-r
\end{array}\right)=\frac{n!}{(n-r)!r!}$
Thus we have shown that (for $0\leq r\leq n$):
$\displaystyle \left(\begin{array}{l}n\\r\end{array}\right)=
\left(\begin{array}{l}n\\n-r
\end{array}\right)$