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Dec 19, 2017 at 17:07 comment added darij grinberg This may work, but probably not as easily as the proofs in Zeilberger's paper. The comment by André Camargo in terrytao.wordpress.com/2017/08/28/… (from 30 August, 2017 at 7:25 pm) sketches how the $r = n$ case can be obtained from Sylvester's identity (which is combinatorially proven in a paper by Berliner and Brualdi). I suspect the same argument goes through for all $r$, but have not checked.
Dec 19, 2017 at 13:42 history answered Douglas Lind CC BY-SA 3.0