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for diagonal elements we have  2 ^(n) choices now for other elements that are left  n^(2) - n becoz relation is symmetric  u have to divide (n^(2) - n) elements by 2  nw other choices left are 2^(n^(2) - n) therfore total no of choices are  2^(n) * 2^(n^(2) - n) it comes out to be  2^((n*(n+1)) / 2)
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