If the value of n ( Y ) + n ( Z ) is k 2 , then | k | is [2024]

Question

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 $A$ . [2024]

Q. If the value of $n (Y) + n (Z)$ is $k^{2}$ , then $| k |$ is _____.

Smart Achievers · Accepted Answer

(36)

Given $S = {1, 2, 3, 4, 5, 6}$ $R$ $S$

Number of elements in $R = 6$

and for each $(a, b) \in R;$ $| a - b |$ $Y = {R \in X : The range of R has exactly one element}$

$X \to$ set of all such relations $R : S \to S$

If

Total number of ordered pairs $(a, b)$ such that $| a - b |$ $\geq 2 = 20$

$∴$ $n (X)$ = number of elements in $X = {}^{20}C_{6}$

$∴$ $m = 20$

$Y = {R \in X : The range of R has exactly one element} .$

From above, if range of $R$ has exactly one element, then maximum number of elements in $R$ will be $4$

$∴$ $n (Y) = 0$

$Z = {R \in X : R is a function from S to S}$

$n (Z) = {}^{4}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{3}C_{1} \times {}^{4}C_{1} = {(36)}^{2}$

$n (Y) + n (Z) = 0 + {(36)}^{2} = k^{2}$

$\Rightarrow$ $| k |$ $= 36$

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