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}$
$0!=1$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
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Using these definitions,
$\displaystyle \left(\begin{array}{l}
n\\
0
\end{array}\right)=\frac{n!}{0!n!}=\frac{n!}{1\cdot n!}=1$
$\displaystyle \left(\begin{array}{l}
n\\
n
\end{array}\right)=\frac{n!}{n!0!}=\frac{n!}{n!1}=1$
