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QUESTION

Prove

    \begin{description}

     \item[(a)] using the factorial definition,

     \item[(b)] using the word definition
    \end{description}

    \begin{description}

     \item[(i)]$\left( \begin{array}{c} n\\ n-r \end{array}
     \right)=\left( \begin{array}{c} n\\ r \end{array}
     \right)$

     \item[(ii)]$\left( \begin{array}{c} n+1\\ r \end{array}
     \right)=\left( \begin{array}{c} n\\ r \end{array}
     \right)+\left( \begin{array}{c} n\\ r-1 \end{array}
     \right)$
    \end{description}

\bigskip

ANSWER

(i) (a) $\displaystyle\left( \begin{array}{c}n\\ n-r \end{array}
   \right)=\frac{n!}{(n-r)!(n-(n-r))!}=\frac{n!}{(n-r)!r!}=
   \left( \begin{array}{c} n \\ r \end{array}\right)$\\
   (b)$\left( \begin{array}{c} n \\r \end{array}\right)$ is the
   number of ways of choosing $r$ from $n$, each choice of $r$ is
   effectively one different choice of $n-r$ because $n-r$ are left
   behind.\\
   (ii) (a)\begin{eqnarray*}\left( \begin{array}{c} n\\ r \end{array} \right) +
   \left( \begin{array}{c} n\\ r-1 \end{array} \right
   )&=& \frac{n!}{(n-r)!r!}+\frac{n!}{(n-r+1)!(r-1)!}\\&=&\frac{n!}{(n-r+1)!r!}\{n-r+1+r\}\\
   &=&\frac{(n+1)!}{(n+1-r)!r!}=\left( \begin{array}{c} n+1\\r\end{array} \right)
   \end{eqnarray*}\\
   (b) Consider $n+1$ things. $\underbrace{\times\times\times\times\ldots\times}_n
   \underbrace{{\mathrm O}}_1$

   Choosing $r$ from $n+1$ is$\left( \begin{array}{c} n+1\\r\end{array}
   \right)$. Each choice either included O in which case you need
   to choose $r-1$ from $n$, $\left( \begin{array}{c} n\\ r-1 \end{array} \right
   )$, or excludes O, in which case we need to choose $r$ from $n$,
   $\left( \begin{array}{c} n\\ r \end{array} \right)$. Hence
   $$\left( \begin{array}{c} n+1\\r\end{array}
   \right)=\left( \begin{array}{c} n\\ r \end{array} \right) \left( \begin{array}{c} n\\ r-1 \end{array} \right
   )$$


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