Sets and Relations JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Sets and Relations, free to read — no sign-in needed. The full chapter has 7 questions; sign in to attempt the remaining 4 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Let $S = \{1, 2, 3, \ldots, 10\}$. Consider the set $X = \{R : R$ is an equivalence relation on the set $S$ such that $R$ has exactly $42$ elements$\}$. Then the number of elements in $X$ is _____________.
   Show answer & solution

   Answer: 2520

   An equivalence relation on a set $S$ corresponds to a partition of $S$ into disjoint equivalence classes. Let the sizes of these equivalence classes be $a_1, a_2, \ldots, a_k$. The number of elements in the equivalence relation $R$ is given by the sum of the squares of the sizes of its equivalence classes: $\sum_{i=1}^k a_i^2 = 42$ Since the total number of elements in $S$ is $10$, we also have: $\sum_{i=1}^k a_i = 10$ We need to find all possible partitions of $10$ such that the sum of their squares is $42$. Let's check the possible sizes of the largest equivalence class, $a_1$: Case 1: $a_1 = 6$ $a_1^2 = 36$. The remaining sum of squares is $42 - 36 = 6$, and the remaining sum of elements is $10 - 6 = 4$. The only way to partition $4$ such that the sum of squares is $6$ is $2, 1, 1$ (since $2^2 + 1^2 + 1^2 = 6$). Thus, one valid partition is $\{6, 2, 1, 1\}$. Case 2: $a_1 = 5$ $a_1^2 = 25$. The remaining sum of squares is $42 - 25 = 17$, and the remaining sum of elements is $10 - 5 = 5$. The only way to partition $5$ such that the sum of squares is $17$ is $4, 1$ (since $4^2 + 1^2 = 17$). Thus, another valid partition is $\{5, 4, 1\}$. Case 3: $a_1 \le 4$ The maximum possible sum of squares would be for the partition $\{4, 4, 2\}$, which gives $4^2 + 4^2 + 2^2 = 36 \lt 42$. No other partitions can yield a sum of $42$. Now, we calculate the number of ways to form these partitions from the $10$ elements of $S$. For the partition $\{6, 2, 1, 1\}$: The number of ways to divide $10$ elements into groups of sizes $6, 2, 1, 1$ is: $\dfrac{10!}{6! \times 2! \times 1! \times 1! \times 2!} = \dfrac{3628800}{720 \times 2 \times 1 \times 1 \times 2} = \dfrac{3628800}{2880} = 1260$ (Note: We divide by $2!$ because there are two groups of identical size $1$). For the partition $\{5, 4, 1\}$: The number of ways to divide $10$ elements into groups of sizes $5, 4, 1$ is: $\dfrac{10!}{5! \times 4! \times 1!} = \dfrac{3628800}{120 \times 24 \times 1} = \dfrac{3628800}{2880} = 1260$ Total number of equivalence relations in $X$ is: $1260 + 1260 = 2520$ Answer: $2520$
2. Q2JEE Advanced Adv 2017 (Paper 2)
   Let $S=\left\{1,2,3,\ldots .9\right\}$. For $k=1,2,\ldots 5,$ let ${N}_{k}$ be the number of subsets of $S$, each containing five elements out of which exactly $k$ are odd. Then ${N}_{1}+{N}_{2}+{N}_{3}+{N}_{4}+{N}_{5}=$
   1. A.$125$
   2. B.$252$
   3. C.$210$
   4. D.$126$
   Show answer & solution

   Answer: (D)

   ${N}_{1}+{N}_{2}+{N}_{3}+{N}_{4}+{N}_{5}=$ Total ways – {when no odd} Total ways ${=}^{9}{C}_{5}$ Number of ways when no odd, is zero ($∵$ only available even are $2,4,6,8$) $∴{}^{9}{C}_{5}-zero=126$
3. Q3JEE Advanced Adv 2010 (Paper 2)
   Let $S=\{1,2,3,4\}$. The total number of unordered pairs of disjoint subsets of $S$ is equal to
   1. A.25
   2. B.34
   3. C.42
   4. D.41
   Show answer & solution

   Answer: (D)

   Let $A \cap B=\phi, A, B \subset S$ $$\begin{aligned} & 3^4=\frac{81+1}{2}=41 \\ \Rightarrow \quad & \frac{3^4+1}{2}=41 \end{aligned}$$