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

As per us, the correct answer is 3. However, NTA has given the final answer key as 4. If we've missed something, please tell us! This question took 2 hours away, ahah.

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Solution:

We need to find relations on $A = \{a, b, c\}$ that contain $(a,b)$ and $(b,c)$, are reflexive and transitive, but not symmetric.

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Since the relation must be reflexive, it must contain $(a,a)$, $(b,b)$, and $(c,c)$.

We are also told it must contain $(a,b)$ and $(b,c)$.

Now, applying transitivity:

$(a,b) \in R$ and $(b,c) \in R \Rightarrow (a,c) \in R$

No other new pairs are forced by transitivity from these elements.

So the minimum required set is:

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

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The remaining pairs that could optionally be added are:

$(b,a),\ (c,b),\ (c,a)$

This gives us $2^3 = 8$ possible subsets to check. For each, we verify whether adding it to $R_0$ keeps the relation transitive and ensures it is not symmetric.

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Adding no extra pairs — $R_0$ alone:

$(a,b) \in R$ but $(b,a) \notin R$, so it is not symmetric. Transitivity already holds.

This is valid. ✓

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Adding $\{(b,a)\}$:

$(b,a) \in R$ and $(a,b) \in R \Rightarrow (b,b) \in R$ ✓ already present

$(b,a) \in R$ and $(a,c) \in R \Rightarrow (b,c) \in R$ ✓ already present

Transitivity holds. $(b,c) \in R$ but $(c,b) \notin R$, so it is not symmetric.

This is valid. ✓

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Adding $\{(c,b)\}$:

$(c,b) \in R$ and $(b,c) \in R \Rightarrow (c,c) \in R$ ✓ already present

$(c,b) \in R$ and $(b,b) \in R \Rightarrow (c,b) \in R$ ✓

$(a,c) \in R$ and $(c,b) \in R \Rightarrow (a,b) \in R$ ✓ already present

Transitivity holds. $(a,b) \in R$ but $(b,a) \notin R$, so it is not symmetric.

This is valid. ✓

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Adding $\{(c,a)\}$:

$(c,a) \in R$ and $(a,b) \in R \Rightarrow (c,b)$ must be in $R$, but it is not present.

Transitivity fails. ✗

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Adding $\{(b,a),\ (c,b)\}$:

$(c,b) \in R$ and $(b,a) \in R \Rightarrow (c,a)$ must be in $R$, but it is not present.

Transitivity fails. ✗

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Adding $\{(b,a),\ (c,a)\}$:

$(c,a) \in R$ and $(a,b) \in R \Rightarrow (c,b)$ must be in $R$, but it is not present.

Transitivity fails. ✗

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Adding $\{(c,b),\ (c,a)\}$:

$(b,c) \in R$ and $(c,a) \in R \Rightarrow (b,a)$ must be in $R$, but it is not present.

Transitivity fails. ✗

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Adding $\{(b,a),\ (c,b),\ (c,a)\}$ — all three optional pairs:

Transitivity holds (all forced pairs are now present). However, every pair now has its reverse in $R$:

$(a,b) \leftrightarrow (b,a)$, $(b,c) \leftrightarrow (c,b)$, $(a,c) \leftrightarrow (c,a)$

The relation is symmetric, which violates the condition. ✗

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Out of $8$ possible combinations, only $3$ satisfy all conditions: reflexive, transitive, containing $(a,b)$ and $(b,c)$, but not symmetric.

The answer is $\boxed{3}$