I can't get very far with this one :/
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$$\left.\binom{m+n}m\middle/\binom{m+n}{m+1}\right. =\left.\frac{(m+n)!}{m!n!}\middle/\frac{(m+n)!}{(m+1)!(n-1)!}\right.$$
$$=\frac{(m+1)\cdot m! (n-1)!}{m!\cdot n\cdot(n-1)!}$$
robjohn
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lab bhattacharjee
- 274,582
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Are you solving from the left or right of the equal sign? – user135094 Mar 13 '14 at 03:13
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@user135094, neither. I've tried to find the ratio of the two binomial coefficients, then apply $$\frac ab=\frac cd\iff ad=bc$$ – lab bhattacharjee Mar 13 '14 at 03:20
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@user135094: just cancel the numerator and denominator in the last fraction – robjohn Mar 13 '14 at 04:17
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Consider $$ \begin{align} \text{LHS} &=\frac{n(m+n)!}{m!(m+n-m)!}\\ &=\frac{n(m+n)!}{m!n!}\\ &=\frac{(m+n)!}{m!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)m!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)!((m+n)-(m+1))!}\\[9pt] &=\text{RHS} \end{align} $$
0
Using the identity from this answer
$$
r\binom{n}{r}=n\binom{n-1}{r-1}\tag{1}
$$
and the basic
$$
\binom{n}{m}=\binom{n}{n-m}\tag{2}
$$
we get
$$
\begin{align}
(m+1)\binom{n+m}{m+1}
&=(n+m)\binom{n+m-1}{m}\tag{3}\\
&=(n+m)\binom{n+m-1}{n-1}\tag{4}\\
&=n\binom{n+m}{n}\tag{5}
\end{align}
$$
Explanation:
$(3)$: apply $(1)$
$(4)$: apply $(2)$
$(5)$: apply $(1)$
