Let A = a, b, c. Then number of relations containing (a, b) and (b, c…

The correct answer is Option (4) →4

Note: The given answer is as per NTA. However, the correct answer should be Option 3 as explained below.

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

$\text{Reflexive} \Rightarrow (a,a),(b,b),(c,c)$ are included

$\text{Given } (a,b),(b,c)$

$\text{Transitive} \Rightarrow (a,c)$ must be included

$\text{Now base relation: } R_0=\{(a,a),(b,b),(c,c),(a,b),(b,c),(a,c)\}$

$\text{Remaining pairs: } (b,a),(c,b),(c,a)$

$\text{Check transitivity restrictions:}$

$\text{If } (c,b)\in R,\ \text{then } (c,a)\in R \text{ (since } (c,b),(b,a) \Rightarrow (c,a)\text{)}$

$\text{If } (b,a)\in R,\ \text{no new compulsory addition}$

$\text{Possible transitive cases:}$

$1.\ \{\}$

$2.\ \{(b,a)\}$

$3.\ \{(c,a)\}$

$4.\ \{(b,a),(c,a)\}$

$5.\ \{(c,b),(c,a)\}$

$6.\ \{(b,a),(c,b),(c,a)\}$

$\text{Now remove symmetric relations}$

$\text{Symmetric requires: if } (a,b)\Rightarrow(b,a),\ (b,c)\Rightarrow(c,b),\ (a,c)\Rightarrow(c,a)$

$\Rightarrow$ only case 6 is symmetric

$\text{Also case 4 becomes symmetric partially? No, since } (c,b)\notin R$

$\text{Now check "not symmetric":}$

$\text{Valid cases excluding symmetric: } 1,2,3$

$\Rightarrow \text{Total} = 3$

The correct answer is $3$.