I can't get very far with this one :/
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3 Answers
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3
$$\left.\binom{m+n}m\middle/\binom{m+n}{m+1}\right. =\left.\frac{(m+n)!}{m!n!}\middle/\frac{(m+n)!}{(m+1)!(n-1)!}\right.$$
$$=\frac{(m+1)\cdot m! (n-1)!}{m!\cdot n\cdot(n-1)!}$$
answered Mar 13, 2014 at 3:10
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$\begingroup$ Are you solving from the left or right of the equal sign? $\endgroup$ Commented Mar 13, 2014 at 3:13
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$\begingroup$ @user135094, neither. I've tried to find the ratio of the two binomial coefficients, then apply $$\frac ab=\frac cd\iff ad=bc$$ $\endgroup$ Commented Mar 13, 2014 at 3:20
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$\begingroup$ @user135094: just cancel the numerator and denominator in the last fraction $\endgroup$– robjohn ♦Commented Mar 13, 2014 at 4:17
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Consider $$ \begin{align} \text{LHS} &=\frac{n(m+n)!}{m!(m+n-m)!}\\ &=\frac{n(m+n)!}{m!n!}\\ &=\frac{(m+n)!}{m!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)m!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)!(n-1)!}\\ &=\frac{(m+1)(m+n)!}{(m+1)!((m+n)-(m+1))!}\\[9pt] &=\text{RHS} \end{align} $$
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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 $(1)$
