yes 2^((n-1)^2) is the answer :)

consider a row or column of size n, Number of ways it can we filled with 1's
and -1's(such that product is 1) is
sum of all nCi where i = 0,2,4..... (i = no of -1s) and that will be 2^(n-1)
(same is the number when product is -1 )
so now let f(i,j) is the number of ways to fill the matrix of size i,j
then
f(i,j) = 2^(i-2)*2^(j-2)*f(i-1)(j-1)*2
where f(1,1) = 1;

explanation for f(i,j)
matrix of size (i,j) can be broken into four parts = matrix of size(i-1,j-1)
+ jth column of size i-1+ ith row
of size (j-1) + element at[i,j]

so ans is
number of ways matrix [i-1][j-1] can be filled is f(i,j) multiplied with
when both row and col are 1 and element is 1 or both row and col are -1 and
ele is -1

solving the equation for f(n,n) will give 2^((n-1)^2)

@skript
my method is little bit complex........how did u arrived at solution.....is
there a simple way to get to the same answer?


On Thu, Jul 28, 2011 at 12:11 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:

> @Hemlatha
> this is one of the possible solution
>
> the Question is to find Number of such solutions
>
> On Thu, Jul 28, 2011 at 12:09 PM, Hemalatha <
> hemalatha.amru...@googlemail.com> wrote:
>
>> Give all the primary and  secondary diagonal Elements a value -1 and the
>> rest as 1s.
>>
>> -1 1 1 1 1 -1
>> 1 -1 1 1 -1 1
>> 1 1 -1 -1 1 1
>> 1 1 -1 -1  1 1
>> 1 -1 1 1 -1 1
>> -1 1 1 1 1 -1
>>
>>
>> Regards
>> Hemalatha
>>
>>
>> On Thu, Jul 28, 2011 at 11:29 AM, priyanka goel 
>> <priya888g...@gmail.com>wrote:
>>
>>> @ SkRiPt...
>>> can u pl explain ur ans?
>>>
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>
>
> --
> Sunny Aggrawal
> B-Tech IV year,CSI
> Indian Institute Of Technology,Roorkee
>
>


-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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