The number of relations on the set A = {1, 2, 3} containing at most 6 elements including (1, 2), whi

Q: Why can't we just use a formula for the number of transitive relations?

There is no simple closed-form formula for the number of transitive relations on a set of size $n$ with specific constraints. In JEE Advanced, the examiners test your ability to perform systematic, logical counting. Embrace the process—it builds the mathematical intuition required for higher-level combinatorics.

Q: How do I know I haven't missed any combinations?

The key is to categorize by cardinality (the number of elements). By checking size 4, then size 5, then size 6, you create a partition of the problem space. If you follow this structure, you won't miss any. It is the difference between guessing and engineering a solution.

Q: Why is $(2, 1)$ forbidden?

The problem explicitly states the relation is 'not symmetric'. Symmetry requires that if $(a, b) \in R$, then $(b, a) \in R$. Since $(1, 2)$ is in, adding $(2, 1)$ would force symmetry. Therefore, $(2, 1)$ is the 'forbidden fruit' that would invalidate your relation.

Question

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 ________

Sigma Percentile · Accepted Answer

### The Architecture of Logic: Mastering Relations

Welcome, future engineer. Today, we are not just solving a problem; we are building a logical structure. When you look at a set $A = \{1, 2, 3\}$, do not just see numbers. See a playground of possibilities.

A relation on this set is essentially a subset of the Cartesian product $A 	imes A$, which contains $3^2 = 9$ possible ordered pairs. Our goal is to find how many of these subsets satisfy three strict conditions: reflexivity, transitivity, and the absence of symmetry, all while keeping the size at most 6.

### Phase 1: The Foundation of Reflexivity

Every relation starts with a skeleton. The problem demands reflexivity, which is the 'diagonal of life.'

For a relation to be reflexive, every element must relate to itself. This forces our set $R$ to contain the identity pairs: $\{(1, 1), (2, 2), (3, 3)\}$. This is our base; we cannot remove these, only add to them.

### Phase 2: The Constraint of Symmetry

The problem gives us a mandatory pair: $(1, 2) \in R$. But it also gives us a crucial constraint: the relation must NOT be symmetric.

Symmetry is a mirror. If $(1, 2)$ is in, symmetry would demand that $(2, 1)$ also be in. By forbidding symmetry, the problem has handed us a gift: we are strictly forbidden from adding $(2, 1)$.

### Phase 3: The Transitive Dance

Now, we enter the heart of the problem: transitivity. Transitivity is a chain reaction.

Mathematically, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must be in $R$. This is our litmus test. Every time we add a new pair, we must check if this chain reaction is satisfied.

### Phase 4: The Systematic Count

Let us build our relations by size, starting from our base set $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2)\}$.

**Case 1: Size 4**
Our base set $R_0$ has 4 elements. It is vacuously transitive because there is no $(b, c)$ pair to chain with $(1, 2)$. That is our first valid relation.

**Case 2: Size 5**
We add one more pair from the remaining options: $\{(1, 3), (3, 1), (2, 3), (3, 2)\}$.

- If we add $(1, 3)$, the set is $\{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3)\}$. This is transitive.
- If we add $(3, 2)$, the set is $\{(1, 1), (2, 2), (3, 3), (1, 2), (3, 2)\}$. This is also transitive.
- Adding $(2, 3)$ or $(3, 1)$ fails because they would require a 6th element to complete the transitive chain. Thus, we have 2 valid relations here.

**Case 3: Size 6**
Now we add two pairs. We must ensure the transitive property holds:

- Adding $\{(1, 3), (2, 3)\}$ works because $(1, 2)$ and $(2, 3)$ imply $(1, 3)$, which is present.
- Adding $\{(3, 1), (3, 2)\}$ works because $(3, 1)$ and $(1, 2)$ imply $(3, 2)$, which is present.
- Adding $\{(1, 3), (3, 2)\}$ works because $(1, 3)$ and $(3, 2)$ imply $(1, 2)$, which is present.

### Conclusion

We have found 1 relation of size 4, 2 relations of size 5, and 3 relations of size 6.

Summing these up, we get a total of **6 valid relations**. When you break a complex problem into systematic cases, the chaos disappears, and the logic reveals itself.

Animated Solution for Mathematics - Sets and Relations: 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 ________

Enter Numerical Value:

Visualized Solution

Defining the Set and Reflexivity

Adding the Mandatory Pair $(1, 2)$

Case 1: Relation with 4 Elements

Case 2: Adding a 5th Element

Case 3: Adding 6 Elements - Part A

Case 3: Adding 6 Elements - Part B

Final Calculation and Conclusion

The Sigma Insight: Types of Relations

The Architecture of Logic

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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