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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Thu Jun 08, 2006 10:50 pm Post subject:
Re: Simple question about enumeration



javerberg@gmail.com wrote:
Quote:  I have question that doesn't sound very hard, but I spent some time
searching for an answer without any result. If anyone could give me a
hint I would be very thankful.
Assume I have 2 identical red balls, 2 identical yellow balls, and 7
empty boxes. I can then put the balls into the boxes in 7x6x5x4 / (2x2)
= 210 different ways (assuming not more than one ball fit into a single
box).
How can I enumerate these possibilities? In other words I would like to
have a bijection between the set of combinations and the set of natural
numbers between 1 and 210. Once again, thankful for any hint.

Probably the easiest way is to:
(1) Generate all 4combinations of {1,2,...,7}; call this enumeration
E(1),
(2) Generate the 4combinations RRYY, RYRY, RYYR, YRRY, YRYR, YYRR,
call it E(2),
(3) Create a bijection from E(1) x E(2) to the possibilities, in the
obvious way.
For instance,
f({1,3,5,6}, YRRY) = Yellows in 1, 6, Reds in 3,5.
 Christopher Heckman 

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javerberg@gmail.com science forum beginner
Joined: 08 Jun 2006
Posts: 1

Posted: Thu Jun 08, 2006 8:15 pm Post subject:
Simple question about enumeration



I have question that doesn't sound very hard, but I spent some time
searching for an answer without any result. If anyone could give me a
hint I would be very thankful.
Assume I have 2 identical red balls, 2 identical yellow balls, and 7
empty boxes. I can then put the balls into the boxes in 7x6x5x4 / (2x2)
= 210 different ways (assuming not more than one ball fit into a single
box).
How can I enumerate these possibilities? In other words I would like to
have a bijection between the set of combinations and the set of natural
numbers between 1 and 210. Once again, thankful for any hint.
Kind regards
Magnus Jäverberg 

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