JEE Advanced 2024 Paper 2 · Q15 · Permutations & Combinations
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 $|a - b| \ge 2$. Let $Y = \{R \in X : \text{The range of } R \text{ has exactly one element}\}$ and $Z = \{R \in X : R \text{ is a function from } S \text{ to } S\}$. Let $n(A)$ denote the number of elements in a set $A$.
If the value of $n(Y) + n(Z)$ is $k^{2}$, then $|k|$ is ______.
Reveal answer + step-by-step solution
Correct answer:36.00
Solution
$Y$: $R$ has range a single element. Then all $6$ pairs in $R$ have the same second coordinate $b$; the first coordinates are $6$ elements of $S$, but with $|a-b|\ge 2$. From the count in Q14, no $b \in S$ has $\ge 6$ valid $a$'s (max is $4$). Hence $n(Y) = 0$. $Z$: $R$ is a function $S \to S$ with $|f(a) - a| \ge 2$ for all $a \in S$. Allowed images: $a=1 \to \{3,4,5,6\}$ (4 choices); $a=2 \to \{4,5,6\}$ (3); $a=3 \to \{1,5,6\}$ (3); $a=4 \to \{1,2,6\}$ (3); $a=5 \to \{1,2,3\}$ (3); $a=6 \to \{1,2,3,4\}$ (4). $n(Z) = 4 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 4 = 1296$. $n(Y) + n(Z) = 1296 = 36^{2}$, so $|k| = 36$.
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