Some kind of Counting problem

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fatemeh8989

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hi, need help about this:
How many matrixes (m*n) there are with 0 & 1 entries such that the number of 1s in every row and every column be an odd number.
thank you
 
As always, what are your thoughts?

If you have no thoughts, you could start drawing some examples until you have some thoughts.

Then, come back here and share them. Cheers :cool:
 
mmm4444bot said:
As always, what are your thoughts?

If you have no thoughts, you could start drawing some examples until you have some thoughts.

Then, come back here and share them. Cheers :cool:
Click to expand...

yes... i already did that... i thought about with special cases like [A](m*n) which m and n both be even or odd.
ummmm in fact I can find some rules for small matrixes but have no idea about general case.
if number of rows be even... so we know that in a specified row should be at least one 1.(lets build this matrix)if I want to change any zero in that row to 1, i have to do it for another 0 too, ( to keep the 1s odd)... we can move on this way...
but in this way we have to care about building columns as we go through...
if number of rows be odd we know that we cant have one, or 3 or odd number of zero...
almost thats all that i can say for this moment... any help..?
 
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fatemeh8989 said:
hi, need help about this:
How many matrixes (m*n) there are with 0 & 1 entries such that the number of 1s in every row and every column be an odd number.
thank you
Click to expand...

If all the entries of matrix is 1, the matrix will be linearly dependent. Would such a matrix be admissible in this case?

If such matrix is admissible, and since there is no restriction on m or n, we can have infinitely many of such matrices whenever both m and n are odd.
 
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I find out the solution I think...
Ummm... as an Idea we can think this way...
let have A[aij] (m-1 * n-1 ) any matrix with entries with o and 1. how many matrixes there are so..? any entry can be 1 or 0...
so we have 2^(m-1)(n-1) such matrixes.
now we add last row and column... but... if the number of 1s in m-1 entries in a row be odd we put last entry 0, else put it 1.
so last row and column will fill this way that will fulfill the situation... but this adding dont add any new case to our previous existing matrixes...
it is almost done... but there was some important points... adding 1 or 0 will not disturb the situation of rows and columns that have been made before.. IF : if m be odd then n be odd... or if m be even then n be even too.
the point I was doubting before i submit this,,.. was that is there any other matrixes that : m be odd and n be even or n be odd and m be even...
i dont have any proof for this but can say that such a matrix can't be found in these conditions.. ( rows and columns with of number odd 1 )
so the number of such a matrixes is 2^(m-1)(n-1)...
I will be approciated for any better idea.
thanks :)
 
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