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mathmath333
 one year ago
Prove
mathmath333
 one year ago
Prove

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & r\times \binom{n}{r}=n\times \binom{n1}{r1},\ \ n\geq r\hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is this true by the way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0write 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3hint: we have the subsequent steps: \[\large \begin{gathered} r \cdot \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) = r\frac{{n!}}{{r!\left( {n  r} \right)!}} = r \cdot \frac{{\left( {n  1} \right)!n}}{{\left( {r  1} \right)!r\left( {n  r} \right)!}} = \frac{{\left( {n  1} \right)!n}}{{\left( {r  1} \right)!\left( {n  r} \right)!}} \hfill \\ \hfill \\ \left( {\begin{array}{*{20}{c}} {n  1} \\ {r  1} \end{array}} \right) = \frac{{\left( {n  1} \right)!}}{{\left( {r  1} \right)!\left( {n  1  r + 1} \right)!}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3\[\begin{gathered} r \cdot \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) = r\frac{{n!}}{{r!\left( {n  r} \right)!}} = r \cdot \frac{{\left( {n  1} \right)!n}}{{\left( {r  1} \right)!r\left( {n  r} \right)!}} = \frac{{\left( {n  1} \right)!n}}{{\left( {r  1} \right)!\left( {n  r} \right)!}} \hfill \\ \hfill \\ \left( {\begin{array}{*{20}{c}} {n  1} \\ {r  1} \end{array}} \right) = \frac{{\left( {n  1} \right)!}}{{\left( {r  1} \right)!\left( {n  1  r + 1} \right)!}} \hfill \\ \end{gathered} \]
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