NIMCET 2026 — Mathematics PYQ

To solve this, we analyze the condition for each element $x\in \{1,\dots ,n\}$ independently. For any element $x$, it can either be in set $A$ or not, in set $B$ or not, and in set $C$ or not.

Let $x_A,x_B,x_C$ be indicator variables that are $1$ if $x$ is in the respective set, and $0$ otherwise. The condition $(A\cap B)\subseteq C\subseteq (A\cup B)$ must hold for each element $x$.

We can list the possible membership combinations for an element $x$ in sets $A,B$, and $C$:

Counting the "Yes" rows, we find that for each element $x$, there are exactly $6$ valid assignments of membership. Since there are $n$ elements and each element can be assigned independently in these $6$ ways, the total number of such triples is $6^n$.

Correct Option: 2