Binomial Iteration

[sowmitra]

Prove that,
\[\displaystyle\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}\forall n\in\mathbb{N}\]

[SANZEED]

$\binom{\binom{n}{2}}{2}=\frac{\frac{n(n-1)}{2}\times (\frac{n(n-1)}{2}-1)}{2}=\frac{n(n-1)(n^{2}-n-2)}{2\times 4}=\frac{n(n-1)(n-2)(n+1)}{8}$.
On the other hand,$3\binom{n+1}{4}=3\times \frac{(n+1)n(n-1)(n-2)}{4!}=\frac{n(n-1)(n-2)(n+1)}{8}$.
The result follows.

[sowmitra]

@ Sanzeed: Wow, man, you are getting really good at providing hard-core analytic solutions. My solution was a Combinatorial one, But, your one works just as well.

[SANZEED]

Thanks @Sowmitra vai. Actually I didn't thought of a combinatorial solution because I find it a lot easier for me to use analytic solution than a combinatorial one. And here is another one I found out from the Principles and Techniques in Combinatorics .
We know that $(x+y)^{n}=\displaystyle\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^{r}$. Now let $x=1$ then,
$(1+y)^{n}=\displaystyle\sum_{r=0}^{n}\binom{n}{r}y^{r}$. Taking the second derivative,
$n(n-1)(1+y)^{n-2}=\displaystyle\sum_{r=0}^{n}r(r-1)\binom{n}{r}y^{r-2}$. Now letting $y=1$ we have,
$n(n-1)2^{n-2}=\displaystyle\sum_{r=0}^{n}(r^{2}-r)\binom{n}{r}=\displaystyle\sum_{r=0}^{n}r^{2}\binom{n}{r}-\displaystyle\sum_{r=0}^{n}r\binom{n}{r}=\displaystyle\sum_{r=0}^{n}r^{2}\binom{n}{r}-n2^{n-1}$.
That is, $\displaystyle\sum_{r=0}^{n}r^{2}\binom{n}{r}=n(n-1)2^{n-2}+n2^{n-1}=n2^{n-2}(n-1+2)=n(n+1)2^{n-2}$.

[sowmitra]

@Sanzeed: তুমিও 'Principles and Techniques in Combinatorics' থেকে পড়। Awesome book.......!!! সিঙ্গাপুরিয়ানরা আসলেই কম্বিতে খুব ভাল।