LET A={123},THEN THE NO OF RELATIONS CONTAING (!,2),(2,3) WHICH ARE REFLEXIVE AND TRANSITIVE BUT NOT SYMMETRIC ARE.GUYS PLS EXPLAIN BRIEFLY.....................
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39Possible number of ordered pairs = 9.
They are (1,2), (1,3), (1,1), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3).
Now each subset should have (1,2) and (2,3) and satisfy the conditions : (a,a) belongs to R and while (a,b) belongs to R, (b,c) belongs to R then (a,c) belongs to R.
The first two ordered pairs, (1,2) and (2,3) are required in every subset and are arranged in 2 ways.
The next three ordered pairs are (1,1), (2,2), (3,3). They are arranged in 3! ways. This wraps up the reflexive condition. Now for the transitive condition, as (1,2) and (2,3) are present, (1,3) will also have to be present. (1,3) can be arranged in only one way. No other ordered pairs satisfy this condition.
So total number of ways = 2 x 3! x 1 = 12
Now the order of arrangements can be reversed and permutations can be done again. So total ways = 12 x 2 = 24.
Each different arrangement or way constitutes a subset or relation. So there are 24 possible relations.
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