The number of relations on the set A={1,2,3}, containing at most 6 elements including (1,2), which are reflexive and transitive but not symmetric, is __________.

Solution

Maths sir · 4 steps

Reflexive plus (1,2) already fixes 4 pairs, so you only choose up to 2 extra pairs and test each set.

1. Fixed pairs. Reflexive forces (1,1), (2,2), (3,3) in, and (1,2) is given. That is 4 elements already, so at most 2 more can be added from (2,1), (1,3), (3,1), (2,3), (3,2).
2. Rules for extras. (2,1) can never be added, because with (1,2) it makes the relation symmetric or breaks transitivity. For every other choice, check each chain (a,b), (b,c) and confirm (a,c) is present.

   Extra pairs	Size	Transitivity check
   none	4	no chains to test
   (1,3)	5	no chain needs a missing pair
   (3,2)	5	no chain needs a missing pair
   (1,3), (2,3)	6	(1,2), (2,3) needs (1,3): present
   (1,3), (3,2)	6	(1,3), (3,2) needs (1,2): present
   (3,1), (3,2)	6	(3,1), (1,2) needs (3,2): present

3. Rejected sets. Adding (3,1) alone needs (3,2), and adding (2,3) alone needs (1,3), so both fail. The other two-pair sets each create a chain whose result is missing. None of the six valid sets is symmetric, because (2,1) is never present.
4. So. There are 1 + 2 + 3 = 6 such relations.

The common slip: 3. Counting only the relations with exactly 6 elements gives 3. The 4-element and 5-element relations also have at most 6 elements, so they count too.

The idea to keep. First lock in the pairs the conditions force. Then list the few free choices and test transitivity chain by chain. A small, careful list beats any shortcut formula here.

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