I agree with skript: Number of ways of doing this is n! One in the first row can be placed in n ways. After one in first row has been placed, we can place One in second row in n-1 ways and so on. So total num of ways is n*(n-1)*...*1 = n!
One possible solution to this problem can be coded as follows:
Input: integer n
Output: different matrices satisfying given criteria
Create an array of integers 1 to n.
do
{
// Omega( n! )
For (i =0 to n-1)
// Omega(n)
print binary representation of (1 << arr[i] ) upto n bits //
Omega(n)
} while( Next_Permutation( arr, arr+n ) );
TC: Omega ( n! ) * Omega ( n ) * omega ( n )
SC: O(n)...( I dunno [?] whether Next_permutation uses any extra space or
what, so i am not considering it... [?] )
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