mathmath333
  • mathmath333
Prove
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & r\times \binom{n}{r}=n\times \binom{n-1}{r-1},\ \ n\geq r\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
is this true by the way
anonymous
  • anonymous
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

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Michele_Laino
  • Michele_Laino
hint: we have the subsequent steps: \[\large \begin{gathered} r \cdot \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) = r\frac{{n!}}{{r!\left( {n - r} \right)!}} = r \cdot \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!r\left( {n - r} \right)!}} = \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}} \hfill \\ \hfill \\ \left( {\begin{array}{*{20}{c}} {n - 1} \\ {r - 1} \end{array}} \right) = \frac{{\left( {n - 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - 1 - r + 1} \right)!}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\begin{gathered} r \cdot \left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) = r\frac{{n!}}{{r!\left( {n - r} \right)!}} = r \cdot \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!r\left( {n - r} \right)!}} = \frac{{\left( {n - 1} \right)!n}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}} \hfill \\ \hfill \\ \left( {\begin{array}{*{20}{c}} {n - 1} \\ {r - 1} \end{array}} \right) = \frac{{\left( {n - 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - 1 - r + 1} \right)!}} \hfill \\ \end{gathered} \]
mathmath333
  • mathmath333
thnks

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