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

The correct answer is Option (4) → 4

Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect, and the correct answer should be Option (3), as explained below.

A reflexive relation on A must contain (a, a), (b, b) and (c, c).

It must also contain (a, b) and (b, c), and transitivity then forces (a, c).

So every such relation contains

    \( R_0 = \{(a,a),(b,b),(c,c),(a,b),(b,c),(a,c)\} \)

The only other pairs are (b, a), (c, b) and (c, a). Checking every way of adding them to \(R_0\):

    Nothing added: \(R_0\) is transitive, and not symmetric since \((b,a) \notin R_0\). Counts.

    (b, a) only: transitive, and not symmetric since \((c,b)\) is missing. Counts.

    (c, b) only: transitive, and not symmetric since \((b,a)\) is missing. Counts.

    (c, a) only: \((c,a),(a,b)\) need \((c,b)\). Not transitive.

    (b, a) and (c, b): \((c,b),(b,a)\) need \((c,a)\). Not transitive.

    (b, a) and (c, a): \((c,a),(a,b)\) need \((c,b)\). Not transitive.

    (c, b) and (c, a): \((b,c),(c,a)\) need \((b,a)\). Not transitive.

    All three: this is \(A \times A\), which is symmetric. Excluded.

So exactly 3 relations qualify.

Option (3) 3 — Should be correct. The three relations are \(R_0\), \(R_0 \cup \{(b,a)\}\) and \(R_0 \cup \{(c,b)\}\).

Option (4) 4 — The answer given by NTA, but it seems to be incorrect. Of the 8 possible relations, 4 are not transitive and 1 is symmetric, which leaves only 3.

Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.