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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Sets and Relations

JEE Advanced 2024 · Paper 2 · Mathematics · Sets and Relations

If the value of n(Y)+n(Z) is k^2 , then |k| is Let (S=\...

If the value of n(Y)+n(Z)n(Y)+n(Z) is k2k^2, then ∣k∣|k| is Let (S=\{1,2,3,4,5,6\}) and XX be the set of all relations RR from SS to SS that satisfy both the following properties: i. RR has exactly 6 elements. ii. For each ((a, b) \in R), we have (|a-b| \geq 2). Let (Y=\{R \in X) : The range of RR has exactly one element (\}) and (Z=\{R \in X: R) is a function from SS to (S\}). Let n(A)n(A) denote the number of elements in a set AA.If the value of n(Y)+n(Z) is k^2 , then |k| is Let (S=\ 1,2,3,4,5,6\ ) and X be the set of all relations R from S to S

Correct answer

0

Step-by-step solution

given ∣a−b∣≥2|a-b| \geq 2 so if If the value of n(Y)+n(Z) is k^2 , then |k| is Let (S=\ 1,2,3,4,5,6\ ) and X be the set of all relations R from S to S i.e. Total elements in X\mathrm{X} is 20C6{ }^{20} \mathrm{C}_6 Now for n(Y)\mathrm{n}(\mathrm{Y}), range of R\mathrm{R} has exactly one element i.e. second elements must be constant in R\mathrm{R} and since R\mathrm{R} must have 6 element so it is not possible to satisfy both condition so n(Y)=0\mathrm{n}(\mathrm{Y})=0.  for  n(z)1→3,4,5,62→4,5,63→1,5,64→1,2,65→1,2,36→1,2,3,4\begin{aligned} & \text{\text{ for } } \quad n(z) \\ & 1 \rightarrow 3,4,5,6 \\ & 2 \rightarrow 4,5,6 \\ & 3 \rightarrow 1,5,6 \\ & 4 \rightarrow 1,2,6 \\ & 5 \rightarrow 1,2,3 \\ & 6 \rightarrow 1,2,3,4 \\ & \end{aligned} no. of relation that are function will be =4C1×3C1×3C1×3C1×3C1×4C1\quad={ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1=(4×3×3)2=k2 i.e. k=36\begin{aligned} & =(4 \times 3 \times 3)^2=k^2 \\ & \text{ i.e. } \mathrm{k}=36\end{aligned}$$
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