Re: identity involving symmetric group

>Let j>=1 be an integer. Let n_1,n_2,...,n_j be j positive integers with
>the following two properties:
>
>1) the average of the n_i is 2
>2) none of the n_i are 2
>
>[e.g. j=5 and the integers could be 1,1,1,1,6]
>
>Consider the product
>
>j(j-n_1)(j-n_1-n_2)(j-n_1-n_2-n_3)...(j-n_1-n_2-...-n_j)
>
>(note that the last term in the product is just -j)
>
>Well, this product is unfair because it relies on the order of the n_i.
>So now sum this product over all j! permutations of the n's.
>
>[i.e. form sum_{g in Symm(j)} j(j-n_{g1})(j-n_{g1}-n_{g2})...(-j) ]
>
>Is this sum always zero?

I find it a bit unnerving that no-one has replied---questions like
this usually get eaten for breakfast in this forum. I take some
heart in the fact that a substantial percentage of our readers are
currently eating turkey, but am a little scared that someone will
ultimately find a counterexample.

As a result of my posting here I have had several replies by email,
all unfortunately telling me things I already knew (I have thought
about the problem a fair bit myself); to stem the flow a bit I
thought I'd post some of these things here.

1) if j is even then it's trivial (the orderings m_1,m_2,...,m_j
and m_j,m_{j-1},...,m_2,m_1 cancel each other out)

2) if all but one of the n_i are 1 then the result boils down
to "alternating sum of binomial coefficients is zero",
or more precisely to the fact that (1-1)^(j-1)=0.

3) I cannot do the case where all but two of the n_i are 1.
Robin Chapman tells me that he has verified the case 1,1,1,1,...,1,a,b
by computer for 3 <= a <= b < 50 (and the sum was always zero)
but the combinatorics of the general case still elude me.
I still cannot say that I "understand" why it works for 1,1,1,3,4,
but it does.

4) Robin also tells me that he has checked the case 1,1,1,...,1,a,b,c
by computer for 3 <= a <= b <= c <= 20 (and the sum was always zero).
This makes me more optimistic that the sum really always is zero.

Kevin Buzzard