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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Feb 2008 11:08:49 IST
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Find number of ways in which one can place the numbers 1, 2, .....n2 on the nxn chessboard, one on each such that numbers in each row and column are in
a)AP b)GP.
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22 Feb 2008 11:39:48 IST
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plz give the answer too
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22 Feb 2008 11:46:49 IST
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Is the answer to the A.P part 8???
Too much for me!! 
I am sorry if its wrong.Extremely difficult 
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__________________________________________________________________________________________________________
From J.R.R. Tolkien's 'The Lord of the Rings':
All that is gold does not glitter
Not all who wander are lost
The old that is strong does not wither,
Deep roots are not reached by frost.
From ashes a fire shall be woken
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king. |
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22 Feb 2008 11:49:00 IST
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yep its seems to be toooo difficult for me too
so @srikant_IIT
plz give the ans
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22 Feb 2008 12:43:16 IST
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hmmm........ let me try 4 ap 1 and n^2 must be at corners as they either begin or end a series.......... now if we apply ssame logic to others......... near 1 ( 2 plces either side or down) we can either place 2 or 3... and not anythng else or placing thm any where else ud brk up series............. and thn....continue puttin adjacent no.on again after in sam way........... well just tell me if there are no faults in this methud thn i will thnk of proceedin 4 calc.
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23 Feb 2008 01:18:05 IST
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For the ap:
Consider the chessboard to be placed on the coordinate axis such that one corner is at the origin. Let  denote the number at position  . Now, must be in a corner, otherwise
Let  Then  must also be in a corner or adjacent to , but for any board larger than 2x2, can't be in  or  because there are no numbers between 1 and 2 to fill the row or column.
And again for any board larger than 2x2, can't be in  because then:
Since is not in a corner, it is adjacent to . So we must have the first (or last) row or column
By the same logic,  would have to be in a corner, and  adjacent to , and there must be a bordering row or column 
Thus every row must be filled with an arithmetic progression of difference , and the columns would therefore have arithmetic progressions of  , or vice versa. 
So there are eight possibilities if you count rotation and flipping.
(for eg, ,  ,  and so on...) Still working on the G.P... This is really a difficult question. What is the source?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
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23 Feb 2008 20:20:14 IST
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Is the answer correct?
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
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23 Feb 2008 20:24:49 IST
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i think GP answer wont exist....because to put all nos on the board there will be prime nos which cannot be obtained with simple GPs...
like 11,17 wont fit in a GP
so answer should be 0
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24 Feb 2008 09:58:02 IST
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yeah, I was thinking along the same lines. 1 and should be placed at the corners of the board (refer my post above). Now let the common ratio of the G.P be  . Since, we are dealing with only natural numbers,  and  .
Let  (the minimum possible value of )
Hence It is not possible the arrange the numbers in a G.P. |
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
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25 Feb 2008 22:46:41 IST
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Only if i could have expressed it like this guy!!!!
Beautiful way.....and i also think it is not possible to arrange them in G.P
Great excellent Konichiwa!!
And Srikant please spill the beans now........about the source of the problem!
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__________________________________________________________________________________________________________
From J.R.R. Tolkien's 'The Lord of the Rings':
All that is gold does not glitter
Not all who wander are lost
The old that is strong does not wither,
Deep roots are not reached by frost.
From ashes a fire shall be woken
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king. |
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8 Mar 2008 20:47:06 IST
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hii
well well .. gud answers .. well done folks ..
btwn .. very good questions also ..
cheers
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Puneet Agrawal
IIT Delhi
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