How many antisymmetric relations are there on a set with n elements ?
(A) 2n.3n(n-1)/2
(B) 2n
(C) n2
(D) n
Answer: (A)
Explanation: Any subset of diagonal pairs is an antisymmetric relation. In a antisymmetric relation each diagonal pair can appear in 2 ways. In antisymmetric relation each non diagonal combination can appear in 3 ways.
By Product rule,
Number of antisymmetric relations possible on A = 2^n. 3^(n(n-1)/2).
Alternate Explanation –
Anti symmetric relation : If R(a,b) and R(b,a) then a=b
Consider all pairs of elements(let us call the elements a,b) among n elements such that a!=b.
In a,b we can form 2 relations (i.e R(a,b) or R(b,a)). We can either keep one of them or none of them but not both. So, we have 3 options. Total number of such a,b pairs = C(n,2) = n*(n-1)/2. So, total options till now = 3^(n(n-1)/2).
Considering the element pairs where a=b, we can either keep them or discard them. There are n such pairs, so total options = 2^n.
So, total options = 2^n. 3^(n(n-1)/2).
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