Let $A = \{a, b, c\}$. Then number of relations containing (a, b) and (b, c) which are reflexive and transitive but not symmetric is

4

The correct answer is Option (4) → 4

$A=\{a,b,c\}$

Given that relation must contain $(a,b)$ and $(b,c)$.

Since relation is reflexive, it must contain

$(a,a),(b,b),(c,c)$.

By transitivity, from $(a,b)$ and $(b,c)$, it must also contain

$(a,c)$.

So compulsory set is

$\{(a,a),(b,b),(c,c),(a,b),(b,c),(a,c)\}$.

Remaining possible pairs are

$(b,a),(c,b),(c,a)$.

For transitivity, these three pairs are independent and can be chosen freely.

Total choices $=2^3=8$.

Relation must not be symmetric. Symmetric occurs only when all three of

$(b,a),(c,b),(c,a)$ are included.

So symmetric cases $=1$.

Required number of relations

$=8-4$

$=4$