Mindyard | Let S = { 1 , 2 , 3 , 4 , 5 , 6 } S = \{1,2,3,4,5,6\} S = { 1 , 2 , 3 , 4 , 5 , 6 } and X X X be the set of all relations R R R from S S S to S S S that satisfy both the following properties: i. R R R has exactly 6 elements. ii. For each ( a , b ) ∈ R (a,b) \in R ( a , b ) ∈ R , we have ∣ a − b ∣ ≥ 2 |a-b| \ge 2 ∣ a − b ∣ ≥ 2 . Let Y=\{R\in X:\text{ The range of }R\text{ has exactly one element}\} Y = { R ∈ X :  The range of  R  has exactly one element } and Z=\{R\in X:R\text{ is a function from }S\text{ to }S\}. Z = { R ∈ X : R  is a function from  S  to  S } . Let n ( A ) n(A) n ( A ) denote the number of elements in a set A A A . 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 $S$ 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 k², then ∣k∣ is ______.

Correct Answer :

Solution :

The correct answer is 36.

Step 1: Understand the Given Sets and Conditions
We are given the set $S = {1, 2, 3, 4, 5, 6}$ .
A relation $R$ from $S$ to $S$ is a subset of $S \times S$ satisfying two conditions:

1. $R$ contains exactly 6 ordered pairs, i.e., $| R | = 6$ .
2. For each ordered pair $(a, b) \in R$ , the condition $| a - b | \geq 2$ must hold.

Step 2: Determine the Number of Elements in Set Y, i.e., n(Y)
Set $Y$ consists of relations $R \in X$ whose range has exactly one element.
Suppose the range of $R$ is ${b_{0}}$ for some fixed $b_{0} \in S$ .
Since $R$ contains 6 distinct ordered pairs and all of them must have the same second element $b_{0}$ , the first elements must be all 6 elements of $S$ .
This means for every element $a \in S$ , we must have $| a - b_{0} | \geq 2$ .

Let's check if any such $b_{0}$ exists:

- If $b_{0} = 1$ , valid choices for $a$ are ${3, 4, 5, 6}$ (only 4 elements available).
- If $b_{0} = 2$ , valid choices for $a$ are ${4, 5, 6}$ (only 3 elements available).
- If $b_{0} = 3$ , valid choices for $a$ are ${1, 5, 6}$ (only 3 elements available).
- If $b_{0} = 4$ , valid choices for $a$ are ${1, 2, 6}$ (only 3 elements available).
- If $b_{0} = 5$ , valid choices for $a$ are ${1, 2, 3}$ (only 3 elements available).
- If $b_{0} = 6$ , valid choices for $a$ are ${1, 2, 3, 4}$ (only 4 elements available).

Since no single value of $b_{0}$ can pair with all 6 elements of $S$ , no such relation exists.
Therefore, $n (Y) = 0$ .

Step 3: Determine the Number of Elements in Set Z, i.e., n(Z)
Set $Z$ consists of relations $R \in X$ that form a valid function from $S$ to $S$ .
Since $| S | = 6$ and $| R | = 6$ , for $R$ to be a function, every element $a \in S$ must be mapped to exactly one element $b \in S$ such that $| a - b | \geq 2$ .

For each element $a \in S$ , let us list the valid choices for $b$ :

- For $a = 1$ : $b \in {3, 4, 5, 6}$ (4 choices)
- For $a = 2$ : $b \in {4, 5, 6}$ (3 choices)
- For $a = 3$ : $b \in {1, 5, 6}$ (3 choices)
- For $a = 4$ : $b \in {1, 2, 6}$ (3 choices)
- For $a = 5$ : $b \in {1, 2, 3}$ (3 choices)
- For $a = 6$ : $b \in {1, 2, 3, 4}$ (4 choices)