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

### Related Formula For a set with n$n$ elements, the number of relations that are both reflexive and symmetric is given by: 2^fracn(n-1)2$$2^{\frac{n(n-1)}{2}}$$ ### Core Logic A reflexive relation must contain all diagonal pairs (x,x)$(x,x)$. There is only 1$1$ way to assign these elements (they MUST be present). A symmetric relation requires that if (x,y)$(x,y)$ is present, (y,x)$(y,x)$ must also be present. Thus, we only have the freedom to choose whether to include the unordered pairs \x, y\$\{x, y\}$ where x neq y$x \neq y$. ### Step 1: Calculate the available independent pairs Number of distinct elements n = 4$n = 4$. Total number of pairs in the cartesian product is n^2 = 16$n^2 = 16$. Number of diagonal pairs (reflexive necessity) = n = 4$= n = 4$. Remaining non-diagonal pairs = 16 - 4 = 12$= 16 - 4 = 12$. Since symmetry pairs them up (a,b) leftrightarrow (b,a)$(a,b) \leftrightarrow (b,a)$, there are exactly frac122 = 6$\frac{12}{2} = 6$ independent choices. ### Step 2: Calculate total relations Each of the 6 independent pairs can either be included or excluded (2$2$ choices). Total relations = 1^4 times 2^6 = 64$= 1^4 \times 2^6 = 64$. ### Pattern Recognition Memorize the combinatorics of binary relations for n$n$ elements: Total = 2^n^2$= 2^{n^2}$, Reflexive = 2^n(n-1)$= 2^{n(n-1)}$, Symmetric = 2^n(n+1)/2$= 2^{n(n+1)/2}$, Reflexive & Symmetric = 2^n(n-1)/2$= 2^{n(n-1)/2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Sets and Relations