Question

Prove

Answer 1

\(\large \color{black}{\begin{align}
& r\times \binom{n}{r}=n\times \binom{n-1}{r-1},\ \ n\geq r\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

is this true by the way

Answer 3

write out the combination nCr using factorials and you'll get a cancellation and then factor out n and rewrite what's left as a combination

Answer 4

hint:
we have the subsequent steps:
\[\large \begin{gathered}
r \cdot \left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right) = r\frac{{n!}}{{r!\left( {n - r} \right)!}} = r \cdot \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!r\left( {n - r} \right)!}} = \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}} \hfill \\
\hfill \\
\left( {\begin{array}{*{20}{c}}
{n - 1} \\
{r - 1}
\end{array}} \right) = \frac{{\left( {n - 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - 1 - r + 1} \right)!}} \hfill \\
\end{gathered} \]

Answer 5

\[\begin{gathered}
r \cdot \left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right) = r\frac{{n!}}{{r!\left( {n - r} \right)!}} = r \cdot \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!r\left( {n - r} \right)!}} = \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}} \hfill \\
\hfill \\
\left( {\begin{array}{*{20}{c}}
{n - 1} \\
{r - 1}
\end{array}} \right) = \frac{{\left( {n - 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - 1 - r + 1} \right)!}} \hfill \\
\end{gathered} \]

Answer 6

thnks