Find the number of ways in which 20 and 20 can be arranged in a row such that upto any point in the row, number of is more than or equal to number of .
Note: You may use calculator, if required.
I posted this question one or two weeks ago (as a problem)! But got no response! So I thought of deleting it and reposting it here as a note. So that some real 'geniuses' can help me solve it out.
Easy Math Editor
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2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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@Michael Mendrin @Calvin Lin @Finn Hulse @Trevor Arashiro @Trevor B. @megh choksi @Sandeep Bhardwaj @Satvik Golechha @Krishna Ar Please help
@John Muradeli @Agnishom Chattopadhyay @Sharky Kesa @Daniel Liu @Christopher Boo @brian charlesworth you too help
This is a fascinating lattices problem. Here is a diagram that might help!
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@Pranjal Jain If you don't understand what I'm talking about, think of ordering the 40 characters like moving on a grid. A step north is like adding an α to the sequence, and a step east is like adding a β.
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1) Pic is not appearing 2) I tried modelling it that way but didnt reached answer.
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n+1(n2n) for n αs and n βs.
Correct!! It comes out to beLog in to reply
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@Pranjal Jain No, a Wiki page does not exist for the Ballot Theorem as yet. Can you add one?
@Finn Hulse Note that to display an image, you have to link to the image file (typically ending with .png, .jpg .gif etc), as opposed to linking to an entire site.
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