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From: lambert@boring.uucp (Lambert Meertens)
Newsgroups: net.math
Subject: Re: Combinatorics question...
Message-ID: <6802@boring.UUCP>
Date: Thu, 27-Feb-86 13:18:44 EST
Article-I.D.: boring.6802
Posted: Thu Feb 27 13:18:44 1986
Date-Received: Sat, 1-Mar-86 17:41:50 EST
References: <736@harvard.UUCP>
Reply-To: lambert@boring.UUCP (Lambert Meertens)
Organization: CWI, Amsterdam
Lines: 29
Apparently-To: rnews@mcvax

In article <736@harvard.UUCP> greg@harvard.UUCP writes:
> How many 8x8 matrices of 0's and 1's are there in which each row and column
> has precisely four 1's?

This reminds me of a problem posed some time ago in this newsgroup: How
many ways are there to deal a deck of cards to four people such that each
person has at least four cards of each suit?

While I do not know how to tackle that one, the 8x8 problem can be solved
by actual enumeration.  A program I wrote returned 116963796250 =
2.5^4.7^2.313.6101.  So, modulo the correctness of the program (which
might well contain bugs since it contains several speed-up tricks), this
is the answer.  Can someone corroborate this?  A more important question:
Have formulas or theories been developed for this type of enumeration?
And: What is the general formula for (2m)x(2m) 0-1 matrices with m 1's in
each row and column?  My program gave:

  F(1) = 2
  F(2) = 90
  F(3) = 297200
  F(4) = 116963796250

The sequence is not in Sloane's handbook.

-- 

     Lambert Meertens
     ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP
     CWI (Centre for Mathematics and Computer Science), Amsterdam