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Best answer
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Number of Anti Symmetric relation is given as below::
|A|=n

|AxA|=n xn

N=Total number of diagonal will n and each one will be 2 option so option (2^n)+non-diagonal element have three option like (3^n^2-n/2)

N=2^n x 3^(n^2-n)/2
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Above number of relation in Anti Symmetric is wrong.

Correct Answer will be ::

N=2^n x 3^(n^2-n)/2

Suppose A={1,2,3}

AxA={(1,1)(1,2)(1,3)

         (2,1)(2,2)(2,3)

         (3,1)(3,2)(3,3)}

Number of diagonal will have 2^n possibility;

Non-diagonl have three possibility whether they can be selected first or selected second or none of them=3^(n^2-n)/2

N=N=2^n x 3^(n^2-n)/2

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we can derive it through the defination of antisymmetric. if (x,y) and (y,x) exist x=y.

so take a set of (1,2,3) now find all relations. of a*a

total 9 ,diagonal elements like (1,1)(2.2)(3,3) should  be included.

total number of choices for them is 2^n  ( i.e. n elemets have 2 choices either has to come or do not appear.)

total number of choices for for lower diagonal elements will be 3.

because either the lower diagonal should come or the upper diagonal elements or non of them comes.

3^(n2-n/2) three choices for these elemnts

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