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From: lambert@boring.UUCP
Newsgroups: net.math
Subject: Re: Combinatorics question...
Message-ID: <6818@boring.UUCP>
Date: Sun, 9-Mar-86 09:25:04 EST
Article-I.D.: boring.6818
Posted: Sun Mar  9 09:25:04 1986
Date-Received: Thu, 13-Mar-86 07:23:29 EST
References: <736@harvard.UUCP> <6802@boring.UUCP> <1184@mmintl.UUCP>
Reply-To: lambert@boring.UUCP (Lambert Meertens)
Organization: CWI, Amsterdam
Lines: 34
Apparently-To: rnews@mcvax

In article <1184@mmintl.UUCP> franka@mmintl.UUCP (Frank Adams) writes:
> In article <6802@boring.UUCP> lambert@boring.UUCP (Lambert Meertens) writes:
>> 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?
>
> I think you must have misquoted the problem.  As stated, the answer is
> simple: zero.  For four people to each have four cards in a suit, there
> must be at least 16 cards in the suit.  But in fact, there are only 13
> cards in each suit.

Indeed.  As Jan Kok pointed out to me, the problem as originally posted
required every person to have at least TWO cards of each suit.

For the problem of the number of (2m)x(2m) 0-1 matrices in which all colums
and rows add up to m, Evangelos Kranakis showed me a simple proof of lower
and upper bounds.  If {2m;m} stands for that number, and (2m;m) for "2m
over m", then

          m                    2m-1
    (2m;m)  <= {2m;m} <= (2m;m)

The upper bound is obtained by considering all ways to fill the first 2m-1
rows so that they add up to m.  The entries in the last row are then
determined by the column-sum requirement.  The lower bound follows from the
fact that after filling the first m rows, one can complete this to a
correct design by copying the complement of the top half into the bottom
half.

-- 

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