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← JEE Advanced 2024 · Paper 2
MathematicsNumerical+3Question 15 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 15

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PARAGRAPH "I"

Let S={1,2,3,4,5,6}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:

  1. RR has exactly 6 elements.
  2. For each (a,b)∈R(a, b) \in R, we have ∣a−b∣≥2|a - b| \ge 2.

Let Y={R∈X:The range of R has exactly one element}Y = \{R \in X : \text{The range of } R \text{ has exactly one element}\} and Z={R∈X:R is a function from S to S}Z = \{R \in X : R \text{ is a function from } S \text{ to } S\}.

Let n(A)n(A) denote the number of elements in a set AA.

If the value of n(Y)+n(Z)n(Y) + n(Z) is k2k^2, then ∣k∣|k| is _____.

Answer and worked solution

Official answer: 35.99

For each value vv, count the a∈Sa\in S with ∣a−v∣≥2|a-v|\ge2:

vv 1 2 3 4 5 6
allowed aa 4 3 3 3 3 4

Y. A relation with range {v}\{v\} and 6 elements would need 66 allowed aa's for that vv. No vv has more than 44, so n(Y)=0n(Y)=0.

Z. A function S→SS\to S in XX assigns each aa a value f(a)f(a) with ∣a−f(a)∣≥2|a-f(a)|\ge2. By symmetry the counts per aa are the same as in the table, so
n(Z)=4⋅3⋅3⋅3⋅3⋅4=1296.n(Z)=4\cdot3\cdot3\cdot3\cdot3\cdot4=1296.

n(Y)+n(Z)=1296=362n(Y)+n(Z)=1296=36^2, so ∣k∣=36|k|=36.