Let A={1,2,3}. The number of relations on A, containing (1,2) and (2,3), which are reflexive and transitive but not symmetric, is _________.

Solution

Maths sir · 3 steps

First write down every pair you are forced to include, then test the few leftover pairs one subset at a time.

1. Forced pairs. Reflexive means (1,1), (2,2) and (3,3) must be there. With (1,2) and (2,3) present, transitivity also forces (1,3). So 6 pairs are fixed.
2. What is left. Only (2,1), (3,2) and (3,1) are still free. That gives 2³ = 8 candidate relations, and each one must be checked for transitivity and for not being symmetric.

   Extra pairs added	Transitive?	Not symmetric?	Counts?
   none	yes	yes, (1,2) without (2,1)	yes
   (2,1)	yes	yes, (1,3) without (3,1)	yes
   (3,2)	yes	yes, (1,3) without (3,1)	yes
   (3,1)	no: (3,1),(1,2) needs (3,2)	not needed	no
   (2,1),(3,2)	no: (3,2),(2,1) needs (3,1)	not needed	no
   (2,1),(3,1)	no: (3,1),(1,2) needs (3,2)	not needed	no
   (3,2),(3,1)	no: (2,3),(3,1) needs (2,1)	not needed	no
   all three	yes	no, this is all of A × A, which is symmetric	no

3. So. Only the first three choices pass every test, so the number of such relations is 3.

The common slip: 4. Forgetting the condition "not symmetric" also lets in the full relation A × A, which gives 4 instead of 3.

The idea to keep. Fix every pair the conditions force before you start counting. Usually only a few pairs stay free, and checking those subsets one by one is quick and safe.

Answer from the NTA final answer key. Solution by PrepWiser.