JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $S = \{1, 2, 3, 4, 5, 6\}$ . Then the number of one-one functions $f: S \to P(S)$ , where $P(S)$ denote the power set of $S$ , such that $f(n) \subset f(m)$ where $n < m$ is

Choose the correct answer:

Explanation

Solution

$f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$

$\text{Let } |f(i)| = k_i, \text{ where } 0 \le k_1 < k_2 < k_3 < k_4 < k_5 < k_6 \le 6$

Case 1: Sizes $\{0, 1, 2, 3, 4, 5\}$ or $\{1, 2, 3, 4, 5, 6\}$

$2 \times (6 \times 5 \times 4 \times 3 \times 2 \times 1) = 1440$

Case 2: One size gap in $\{0, 1, 2, 3, 4, 5, 6\}$

$\text{Gaps possible at index 1, 2, 3, 4, or 5.}$

$5 \times (\text{Ways to choose subsets with one gap of size 2})$

$5 \times (\frac{6!}{2!}) = 5 \times 360 = 1800$

\text{Total} = 1440 + 1800 = 3240