JEE Advanced 2026 · Paper 1, Mathematics question 9

An equivalence relation is a partition of $S$ into classes; a class of size $k$ contributes $k^2$ ordered pairs. So we need partitions of 10 with
$$\sum n_i=10,\qquad\sum n_i^2=42.$$

Possible block sizes. A block of 7 gives $49>42$.

• With a block of 6: the other 4 elements need squares summing to 6, i.e. $2,1,1$. So $(6,2,1,1)$.
• Largest block 5: the other 5 need squares summing to 17, only $4+1$. So $(5,4,1)$.
• Largest block 4: $(4,4,\cdot)$ leaves 2 elements needing 10, impossible; $(4,3,\dots)$ leaves 3 elements needing 17, impossible.

Counting.

• $(6,2,1,1)$: $\dfrac{10!}{6!\,2!\,1!\,1!\,2!}=1260$.
• $(5,4,1)$: $\dfrac{10!}{5!\,4!\,1!}=1260$.

Total $2520$.