AP ICET 2019 Shift 1 Paper

Question : 114 of 200

Marks: +1, -0

If n(4) = 5. then the number of relations on A which are not symmetric is

Solution:

Number of symmetric relations

= 2^{n(n+1){/}2}

Number of relation

= 2^{n^2}

So, the number of relations that are not symmetric

=2^{n^2}-2^{n(n+1){/}2}

Put n =5, we get

2^25 – 2^15 = 2^15(1023)

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