A simple inequality

[Richard Guy]

An interesting challenge, but I doubt if there's
anything neat.  The ratio of the two sides is
about  3 : 2  so one needs to select about 2/3
of a set as being special in some sense.

Algebraically it boils down to  n^2 + 6n > 19
which is true for  n > 3.  With the usual
interpretation, there is equality for  n = 0,
1, 2.        R.

On Tue, 30 Mar 2004 santi_spadaro at virgilio.it wrote:

> There's an old Monthly article (Vol. 87, n.9) of Solomon Golomb which deals
> with iterated binomial coefficients. Among the many elegant results which
> are proved I came to this simple inequality:
> 
> Binomial (Binomial (n,2), 3) > Binomial (Binomial (n,3), 2)
> 
> While providing both an algebraic and combinatorial proof for most of the
> other identities, here he only gives a proof based on numerical estimates
> of the Binomial coefficients. This result also follows trivially from two
> of the reduction formulas which Golomb states at the end of the article,
> namely:
> 
> Binomial (Binomial (n,2),3)= Binomial (n+1,6)+13*Binomial (n+2, 6)+Binomial
> (n+3,6)
> 
> Binomial (Binomial (n,3),2) = 6*Binomial (n+2,6)+3*Binomial (n+1,6)+Binomial
> (n,6)
> 
> I would be very pleased to see a combinatorial proof of the above fact,
> say an injection between 3-multigraphs on 2 edges and n vertices and graphs
> on 3 edges and n vertices. Can you think of any?
> 
> Regards,
> Santino
> 
>