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I have this problem that I am a bit unsure about how to proceed forward with.

Problem: Show that $n{\binom{m+n}{m} = (m+1)\binom{m+n}{m+1}}$ for all integers n, m > 0.

In the solution it says that we should use the definition of binomial coefficient.

Can anyone describe or tell me how to proceed with this problem ?

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    $\begingroup$ First write the definition of the binomial coefficient. $\endgroup$ – Yves Daoust Nov 20 '14 at 14:31
  • $\begingroup$ Is there a typo in your question? $\endgroup$ – hypergeometric Nov 20 '14 at 14:40
  • $\begingroup$ That doesn't seem correct. $n\binom{n+n}{m} = n\frac{2n(2n-1)\dots(2n-m+1)}{m!}$, while $(m+1)\binom{m+n}{m+1} = (m+1)\frac{(m+n)(m+n-1)\dots(m+n-(m+1)+1)}{(m+1)!} = \frac{(m+n)(m+n-1)\dots n}{m!}$. $\endgroup$ – brick Nov 20 '14 at 14:41
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    $\begingroup$ I think $n\binom{n+n}m$ should read $n\binom{m+n}m$ but will await confirmation by OP. $\endgroup$ – hypergeometric Nov 20 '14 at 14:43
  • $\begingroup$ Sorry.. It should be m+n $\endgroup$ – Hanne Nov 20 '14 at 14:48
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$$\begin{align} n\binom{m+n}m &=n\frac{(m+n)!}{m!\ n!}\\ &=\frac{(m+n)!}{m!\ (n-1)!}\\ &=\color{blue}{(m+1)}\frac{(m+n)!}{\color{blue}{(m+1)}\ m!\ (n-1)!}\\ &=(m+1)\frac{(m+n)!}{(m+1)!\ (n-1)!}\\ &=(m+1)\binom{m+n}{m+1}\\ \end{align}$$

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  • $\begingroup$ The third step, how did you get that ? $(m+1)\frac{(m+n)!}{(m+1)!\ (n-1)!}$ $\endgroup$ – Hanne Nov 20 '14 at 14:55
  • $\begingroup$ Is it because $(m+1)! = (m+1)*m*(m-1)*(m-2)....$ ? $\endgroup$ – Hanne Nov 20 '14 at 14:57
  • $\begingroup$ See edited answer. And $(m+1)!=(m+1)m!$. $\endgroup$ – hypergeometric Nov 20 '14 at 14:58
  • $\begingroup$ Ahh..thank you hypergeometric, well explained :) $\endgroup$ – Hanne Nov 20 '14 at 15:00
  • $\begingroup$ You're most welcome :) $\endgroup$ – hypergeometric Nov 20 '14 at 15:00

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