# Definition of binomial coefficient

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 ?

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

\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}
• The third step, how did you get that ? $(m+1)\frac{(m+n)!}{(m+1)!\ (n-1)!}$ – Hanne Nov 20 '14 at 14:55
• Is it because $(m+1)! = (m+1)*m*(m-1)*(m-2)....$ ? – Hanne Nov 20 '14 at 14:57
• See edited answer. And $(m+1)!=(m+1)m!$. – hypergeometric Nov 20 '14 at 14:58