JEE Advanced 2026 Paper 1, Mathematics Section 3 Q1: Relations

An equivalence relation on $S$ corresponds to a partition of $S$ into equivalence classes. If the classes have sizes $n_1,n_2,\dots ,n_k$ with $\sum n_i=10$, the total number of related pairs (the size of the relation) is $\sum n_i^2$.

We need partitions of $10$ for which $\sum n_i^2=42$.

Enumerating partitions of 10 with squared-sum 42:

Partition $\{6,2,1,1\}$: sum $=10$, squared-sum $=36+4+1+1=42$. $✓$

Partition $\{5,4,1\}$: sum $=10$, squared-sum $=25+16+1=42$. $✓$

(Other partitions can be checked to fail this constraint.)

Counting set partitions for each type:

For $\{6,2,1,1\}$ (two singletons of equal size, divide by $2!$):

$\frac{10!}{6!2!1!1!}\cdot \frac{1}{2!}=\frac{10!}{6!\cdot 2!\cdot 2!}=\frac{3628800}{2880}=1260$

For $\{5,4,1\}$ (all distinct block sizes):

$\frac{10!}{5!4!1!}=\frac{3628800}{2880}=1260$

Total: $1260+1260=2520$.