How do we solve this cumulative question involving sequences and binomial coefficients

Question

How is $r*\binom{n}{r}*\binom{n}{r+1}/(r+2)= n*(\binom{n-1}{r-1})*(\binom{n+1}{r+2}/(n+1))$?,

Now this was a term which was a step which was a part of a sequence related question. But my doubt what prompts one to write the RHS directly from LHS? When I equated LHS and RHS, I verified its correct. But how do one get it that the given LHS=RHS.

Answer 1

Here's an approach that avoids factorials and instead uses the absorption identity $k \binom{n}{k} = n\binom{n-1}{k-1}$ twice: $$ \frac{\color{red}{r\binom{n}{r}}\color{blue}{\binom{n}{r+1}}}{\color{blue}{r+2}} = \frac{\color{red}{n\binom{n-1}{r-1}}\color{blue}{\binom{n+1}{r+2}}}{\color{blue}{n+1}}$$

Answer 2

The answer to your question will almost certainly depend on the reason why whoever it was who was making use of the identity found the RHS more convenient for his or her purpose than the LHS. As far as I can tell, neither side of the identity could be considered much simpler than the other, so there seems to me to be no obvious reason why the expression one side of the identity should be more useful in general than that on the other.

However, if you were trying to find a simpler expression for the LHS, here's one fairly natural way you might stumble across the identity. The most obvious thing to do to try and simplify the LHS is to first expand out the binomial coefficients in terms of their component factorials: \begin{align} \frac{r{n\choose r}{n\choose r+1}}{r+2}&=\frac{r(n!)^2}{(r+2)r!(n-r)!(r+1)!(n-r-1)!}\\ &=\frac{(n!)^2}{(r-1)!(n-r)!(r+2)!(n-r-1)!}\ \ . \end{align} If you now notice that the arguments of the first pair of factorials in the denominator, $\ r-1\ $ and $\ n-r\ $, sum to $\ n-1\ $, and that the arguments of the second pair, $\ r+2\ $ and $\ n-r-1\ $, sum to $\ n+1\ $, you could then very easily recognise that \begin{align} \frac{1}{(r-1)!(n-r)!}&=\frac{{n-1\choose r}}{(n-1)!}\ \ \ \ \text{ and}\\ \frac{1}{(r+2)!(n-r-1)!}&=\frac{{n+1\choose r+2}}{(n+1)!}\ , \end{align} which immediately gives you \begin{align} \frac{r{n\choose r}{n\choose r+1}}{r+2}&=\frac{(n!)^2{n-1\choose r}{n+1\choose r+2}}{(n-1)!(n+1)!}\\ &=\frac{n{n-1\choose r}{n+1\choose r+2}}{n+1}\ . \end{align}