Let S = {1, 2, 3, 4, 5, 6} and X be the set o | vedclass

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(A) First,we determine the number of pairs $(a, b)$ such that $a, b \in S$ and $|a-b| \geq 2$.For each $a \in S$,the number of possible $b$ values is:

Vedclass · Accepted Answer

$20, 36$

Vedclass · Answer

(A) First,we determine the number of pairs $(a, b)$ such that $a, b \in S$ and $|a-b| \geq 2$.For each $a \in S$,the number of possible $b$ values is:- If $a=1$,$b \in \{3, 4, 5, 6\}$ ($4$ values)- If $a=2$,$b \in \{4, 5, 6\}$ ($3$ values)- If $a=3$,$b \in \{1, 5, 6\}$ ($3$ values)- If $a=4$,$b \in \{1, 2, 6\}$ ($3$ values)- If $a=5$,$b \in \{1, 2, 3\}$ ($3$ values)- If $a=6$,$b \in \{1, 2, 3, 4\}$ ($4$ values)Total number of such pairs is $4+3+3+3+3+4 = 20$.Since $R$ must have exactly $6$ elements chosen from these $20$ pairs,$n(X) = {}^{20}C_{6}$,so $m = 20$.For $n(Y)$,the range of $R$ has exactly one element,meaning all $6$ pairs $(a, b)$ in $R$ must have the same $b$. However,for any fixed $b$,there are at most $5$ values of $a$ such that $|a-b| \geq 2$. Thus,$n(Y) = 0$.For $n(Z)$,$R$ is a function,meaning for each $a \in S$,there is exactly one $b$ such that $(a, b) \in R$ and $|a-b| \geq 2$. The number of choices for each $a$ is $4, 3, 3, 3, 3, 4$ respectively. Thus,$n(Z) = 4 	imes 3 	imes 3 	imes 3 	imes 3 	imes 4 = 1296 = 36^{2}$.Therefore,$n(Y) + n(Z) = 0 + 36^{2} = 36^{2}$,so $|k| = 36$.

List $I$	List $II$
$P. f_4$ is	$1. \text{onto but not one-one}$
$Q. f_3$ is	$2. \text{neither continuous nor one-one}$
$R. f_2 \circ f_1$ is	$3. \text{differentiable but not one-one}$
$S. f_2$ is	$4. \text{continuous and one-one}$

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