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- Path: sparky!uunet!spool.mu.edu!wupost!usc!news.service.uci.edu!beckman.com!dn66!a_rubin
- Newsgroups: sci.math
- Subject: Re: Combinatorial? problem
- Message-ID: <a_rubin.726176150@dn66>
- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Date: 4 Jan 93 19:35:50 GMT
- References: <Dec.23.18.15.49.1992.22657@pepper.rutgers.edu> <1992Dec29.041921.29638@cbnewsm.cb.att.com>
- Organization: Beckman Instruments, Inc.
- Keywords: binomial coefficients
- Nntp-Posting-Host: dn66.dse.beckman.com
- Lines: 40
-
- In <1992Dec29.041921.29638@cbnewsm.cb.att.com> thf@cbnewsm.cb.att.com (thomas.h.foregger) writes:
-
- >In article <Dec.23.18.15.49.1992.22657@pepper.rutgers.edu>, gore@pepper.rutgers.edu (Bittu) writes:
- >>
- >> Someone please help me with the following problem. I have pretty much
- >> given up on it after a lot of thinking. We want to show the following:
- >>
- >> given m,n nonnegative integers, the quantity
- >>
- >> (2m)! (2n)!
- >> ------------- is an integer.
- >> m! n! (m+n)!
- >>
- >>
- >> Note that this is very easy to show by the standard argument where for
- >> every prime p, you find the highest power of p (say p^k) that divides
- >> the denominator and then show that p^k divides the numerator as well.
- >>
- >> I want a combinatorial proof of this. I have tried rewriting the above
- >> as C(2m,m)*C(2n,n)/C(m+n,m) where C(a,b) is "a choose b" and also in
- >> other ways, but I still haven't come up with a combinatorial proof.
- >>
- >> --Bittu
-
-
- >Without loss of generality assume m<=n so m+n <= 2m.
- ??? >=
- >Write the ratio as
-
- >(2m)! . n! . (2n)!
- >---- --- -----
- >(m+n)! m! n! n!
-
- >Each of the 3 factors is clearly an integer, so the ratio is an integer.
-
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
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