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}$
$0!=1$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!}$
----------------------
Using these definitions,
$\left(\begin{array}{l}
n\\
1
\end{array}\right) =\displaystyle \frac{n!}{1!(n-1)!}$
... note that: $ n(n-1)\cdot...\cdot 2\cdot 1=n(n-1)! ...$
$=\displaystyle \frac{n(n-1)!}{1(n-1)!}=\qquad$reduce $(n-1)!$ ...
$=\displaystyle \frac{n}{1}=n$
$\left(\begin{array}{l}
n\\
n-1
\end{array}\right) =\displaystyle \frac{n!}{(n-1)![n-(n-1)]!}$
... note that $ n(n-1)\cdot...\cdot 2\cdot 1=n(n-1)! ...$
$=\displaystyle \frac{n(n-1)!}{(n-1)!1}=n$.
So,
$\left(\begin{array}{l}
n\\
1
\end{array}\right)=\left(\begin{array}{l}
n\\
n-1
\end{array}\right) =n$.
