If N is the number of antisymmetric relations on a set A=x|x is prime number≤10, then the number of divisors of N is _____

A=2,3,5,7 where n=4N=number of Anti-symmetric relationsN=2n×3n(n−1)2=24×36No. of divisors of N=(4+1)(6+1)=5×7=35

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If N is the number of antisymmetric relations on a set $A = \{x | x is prime number \leq 10\}$ , then the number of divisors of N is _____

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answer is 35.

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$A = \{2, 3, 5, 7\}$ where n=4

N=number of Anti-symmetric relations

$N = 2^{n} \times 3^{\frac{n (n - 1)}{2}} = 2^{4} \times 3^{6}$

No. of divisors of $N = (4 + 1) (6 + 1) = 5 \times 7 = 35$

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