Thread: Problem 2.10
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Old 03-04-2018, 05:48 AM
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magdon magdon is offline
Join Date: Aug 2009
Location: Troy, NY, USA.
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Default Re: Problem 2.10

For example, if you have 10 ways to dichotomize 4 points x_1,\ldots,x_4 and 8 ways to dichotomize another 4 points x_5,\ldots,x_8, for each of the 10 ways for the first 4 points there are at most 8 ways to dichotomize the second 4 points (why?). Therefore, there are at most 10\times 8 ways to dichotomize all 8 points.

Originally Posted by pdsubraa View Post
@Magdon - Can you explain us in detail.

That would help!

Thanks for your time!
Have faith in probability
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