Answer
please see step-by-step
Work Step by Step
(see page 881)
$n!=n(n-1)\cdot...\cdot 2\cdot 1,$ for $ n\in \mathbb{N}$
$0!=1$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}\qquad (*)$
----------------------
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
... multiplication is commutative,
$r!(n-r)!$=$(n-r)!r!$
$=\displaystyle \frac{n!}{(n-r)!r!}$
which equals (*) when r is replaced with (n-r)...
$n-(n-r)=r$
$=\left(\begin{array}{l}
n\\
n-r
\end{array}\right)$ for $0\leq r\leq n$
