Thursday, September 06, 2007
11:06
11:2*3
2*7*79
New approach to the topology/combinatorics problem:
Suppose we have two sets: (x1, x2, ... , xm) and (y1, y2, ... , yn) where m, n are positive integers, and we have a collection P of ordered pairs of the form (xj, yk) where 1 <= j <= m, 1 <= k <=n.
There are m*n possible ordered pairs of this form, and therefore 2^(m*n) possible collections. But suppose we add the following restriction: For any k between 1 and n, there exists j between 1 and m such that (xj, yk) is an element of P. In other words, every element of the second set has to be used in at least one ordered pair.
How many possible collections are there that satisfy this stipulation?
2*7*79
New approach to the topology/combinatorics problem:
Suppose we have two sets: (x1, x2, ... , xm) and (y1, y2, ... , yn) where m, n are positive integers, and we have a collection P of ordered pairs of the form (xj, yk) where 1 <= j <= m, 1 <= k <=n.
There are m*n possible ordered pairs of this form, and therefore 2^(m*n) possible collections. But suppose we add the following restriction: For any k between 1 and n, there exists j between 1 and m such that (xj, yk) is an element of P. In other words, every element of the second set has to be used in at least one ordered pair.
How many possible collections are there that satisfy this stipulation?
# posted by Fibonacci @ 6.9.07 
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I need to find more food; he ate all the things that were in there except the ladybug.
It's not equivalent to the number of topologies, but it's a calculation that could be used in finding them.
It's not equivalent to the number of topologies, but it's a calculation that could be used in finding them.
Found some more Mantis Pictures for you, and a video! If you can, try to get sound for the video. It's quite entertaining.
Clearly your acquisition of a mantis has caused an influx of mantid images on blogs that I frequent.
Yet Another Set
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