Question

The number of relations, defined on the set {a, b, c, d which are both reflexive and symmetric, is equal to:}

• A. 256
• B. 16
• C. 1024
• D. 64

Hint:

Memorize the relation counting formulas for a set of $n$ elements: Total relations = $2^{n^2}$, Reflexive = $2^{n^2-n}$, Symmetric = $2^{\frac{n(n+1)}{2}}$, Both Reflexive and Symmetric = $2^{\frac{n(n-1)}{2}}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the total number of relations on a set with $4$ elements that satisfy both the reflexive and symmetric properties.
Step 2: Key Formula or Approach:
For a set containing $n$ elements, the number of relations that are both reflexive and symmetric is given by the formula $2^{\frac{n(n-1)}{2}}$.
Step 3: Detailed Explanation:
Let the set be $A = \{a, b, c, d\}$. The number of elements in the set is $n = 4$.
A relation $R$ on $A$ is a subset of $A \times A$.
For $R$ to be reflexive, it must contain all diagonal elements: $(a,a), (b,b), (c,c), (d,d)$. Since these must be present, we have $1$ choice for each (they must be included).
For $R$ to be symmetric, if $(x,y) \in R$, then $(y,x)$ must also be in $R$.
Excluding the $n$ diagonal elements, there are $n^2 - n$ non-diagonal elements.
These non-diagonal elements form $\frac{n^2 - n}{2} = \frac{n(n-1)}{2}$ pairs of the form $\{(x,y), (y,x)\}$.
For the relation to remain symmetric, for each of these pairs, we must either include both elements in $R$ or include neither.
This gives us $2$ choices for each of the $\frac{n(n-1)}{2}$ pairs.
Thus, the total number of such relations is $2^{\frac{n(n-1)}{2}}$.
Substitute $n = 4$: \[ \text{Number of relations} = 2^{\frac{4(4-1)}{2}} = 2^{\frac{4 \times 3}{2}} = 2^6 \] \[ 2^6 = 64 \] Step 4: Final Answer:
The number of relations that are both reflexive and symmetric is 64.