Combinatoric Proof: n Choose 2 Choose 2 = 3 (n+1) Choose 4

[anemone]

Show that $${{n\choose 2}\choose 2 }=3 {n+1\choose 4}$$

 

[kaliprasad]

I would expect some one to solve combinatorically but I provide algebragically

Spoiler

writing (n k) for nCk

We have (n 2) = n(n-1)/2

So ((n 2) 2) = ((n(n-1)/2) (n(n-1)/2-1))/2
= n(n-1)/2 ( n^2-n -2)/ 4
= n(n-1)(n-2)(n+1)/8
= 3( n(n-1)(n-2)(n+1)/24
= 3 ( n+1)!/(4!(n+1-4)! Multiply numerator and dnominator by (n-3)!
= 3 (n + 1 4)