For A={1,2,3,4} find number of symmetric relations

Question

If $A = \{1, 2, 3, 4\}$ , find the total number of symmetric relations on $A$ .

Solution — Step by Step

A relation $R$ on a set $A$ is symmetric if: for every $(a, b) \in R$ , we also have $(b, a) \in R$ .

In other words: if the relation includes the ordered pair $(a, b)$ , it must also include $(b, a)$ . If $a = b$ (a diagonal element like $(1, 1)$ ), then the condition is automatically satisfied since $(a, a)$ is its own “pair.”

$|A| = 4$ , so $|A \times A| = 16$ ordered pairs in total.

These 16 pairs split into two categories:

Diagonal pairs (where $a = b$ ): $(1,1), (2,2), (3,3), (4,4)$ — 4 pairs

Off-diagonal pairs (where $a \neq b$ ): The remaining $16 - 4 = 12$ pairs.

These 12 off-diagonal pairs group into 6 symmetric pairs (also called “partner pairs”):

$\{(1,2), (2,1)\}$
$\{(1,3), (3,1)\}$
$\{(1,4), (4,1)\}$
$\{(2,3), (3,2)\}$
$\{(2,4), (4,2)\}$
$\{(3,4), (4,3)\}$

Number of symmetric (unordered) pairs = $\dfrac{n(n-1)}{2} = \dfrac{4 \times 3}{2} = 6$

To build a symmetric relation:

For each diagonal pair $(a, a)$ : We can either include it or exclude it independently. There are 4 diagonal pairs → $2^4$ choices.

For each off-diagonal symmetric pair $\{(a,b), (b,a)\}$ : We must either include BOTH or exclude BOTH (to maintain symmetry). We cannot include just one. There are 6 such pairs → $2^6$ choices.

These choices are independent of each other.

Total symmetric relations = (choices for diagonal pairs) × (choices for off-diagonal symmetric pairs)

= 2^4 \times 2^6 = 2^{4+6} = 2^{10} = 1024

For a set $A$ with $|A| = n$ :

Number of diagonal pairs = $n$
Number of off-diagonal symmetric pairs = $\dfrac{n(n-1)}{2}$
Total choices = $2^n \times 2^{\frac{n(n-1)}{2}} = 2^{n + \frac{n(n-1)}{2}} = 2^{\frac{n(n+1)}{2}}$

For $n = 4$ : $2^{\frac{4 \times 5}{2}} = 2^{10} = 1024$ ✓

Answer: 1024 symmetric relations

Why This Works

The key insight is that symmetry is a “paired constraint” — it doesn’t restrict whether a pair is included, only that partners must travel together. This is why we count independently: each diagonal element is a free choice, and each off-diagonal pair is a single binary choice (both in, or both out).

Compare this with reflexive relations (where all diagonal elements must be included: $2^{n^2 - n}$ choices) or arbitrary relations ( $2^{n^2}$ total subsets of $A \times A$ ).

Alternative Method

Total subsets of $A \times A$ = $2^{16}$ . Of these, we want only those that are symmetric. We use the formula directly:

\text{Symmetric relations on a set of size } n = 2^{\frac{n(n+1)}{2}}

For $n = 4$ : $2^{\frac{4 \times 5}{2}} = 2^{10} = 1024$ .

For CBSE Class 12 and JEE, also know: number of reflexive relations = $2^{n^2 - n}$ ; number of reflexive and symmetric relations = $2^{\frac{n(n-1)}{2}}$ (diagonal elements must all be included, off-diagonal pairs free). These formulas come up as MCQs very frequently.

Common Mistake

The most common error is treating all $n^2$ pairs as independent. Students compute $2^{n^2/2}$ instead of $2^{\frac{n(n+1)}{2}}$ , forgetting that diagonal elements also have a free choice (each diagonal pair is independent of its partner, since it has no separate partner). Always split into diagonal and off-diagonal before counting.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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