JEE Advanced 2024 Paper 2, Mathematics Section 4 Q1: Combinations

JEE Advanced2024Paper 2Mathematics Section 4

PARAGRAPH I

Let $S = {1, 2, 3, 4, 5, 6}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

i. $R$ has exactly 6 elements.

ii. For each $(a, b) \in R$ , we have $∣ a - b ∣ \geq 2$ .

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

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

(There are two questions based on PARAGRAPH “I”, the question given below is one of them)

Q1. If $n (X) =^{m} C_{6}$ , then the value of $m$ is ______.

Your answer

Solution

The total number of ordered pairs $(a, b)$ in $S \times S$ with $∣ a - b ∣ \geq 2$ counts the candidate elements that any relation $R$ can be built from. For each $a \in S$ , list the admissible $b$ 's:

$a = 1$ : $b \in {3, 4, 5, 6}$ — 4 options.

$a = 2$ : $b \in {4, 5, 6}$ — 3 options.

$a = 3$ : $b \in {1, 5, 6}$ — 3 options.

$a = 4$ : $b \in {1, 2, 6}$ — 3 options.

$a = 5$ : $b \in {1, 2, 3}$ — 3 options.

$a = 6$ : $b \in {1, 2, 3, 4}$ — 4 options.

Total admissible pairs $= 4 + 3 + 3 + 3 + 3 + 4 = 20$ .

A relation $R$ with exactly 6 elements is any 6-element subset of these 20 pairs, so $n (X) = (6 20)$ . Therefore $m = 20$ .

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JEE Advanced 2024 Paper 2, Mathematics Section 4 Q1: Combinations

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