JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Sets Relations and Functions

A symmetric relation on a set of $n$ elements is fixed by independently choosing, for each of the $n$ diagonal pairs and each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered off-diagonal pairs, whether it belongs to the relation — that is $n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n+1)}{2}$ free binary choices, giving $2^{n(n+1)/2}$ symmetric relations. For $n=4$, this is $2^{10}=1024$.

If the relation must also be reflexive, all $n$ diagonal pairs are forced in, leaving only the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$ off-diagonal choices free: $2^{n(n-1)/2}$ relations. For $n=4$, this is $2^6=64$.

Symmetric relations that are not reflexive: $1024-64=960$.