Let a set $A = A_1 \cup A_2 \cup ..... \cup A_k$, where $A_i \cap A_j = \phi$ for $i \neq j$, $1 \le j, j \le k$. Define the relation R from A to A by $R = \{(x, y) : y \in A_i$ if and only if $x \in A_i, 1 \le i \le k\}$. Then, R is :

Let $A = \{-2, -1, 0, 1, 2, 3\}$. Let R be a relation on $A$ defined by $xRy$ if and only if $y = \max\{x, 1\}$. Let $l$ be the number of elements in R. Let $m$ and $n$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then $l + m + n$ is equal to

Let $A = {1, 2, 3, 4, 5}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x \le 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m + n$ is equal to :

Let $R = \{(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)\}$ be a relation on the set $A = \{3, 6, 9, 12\}$. The relation is :

An organization awarded $48$ medals in event 'A', $25$ in event 'B' and $18$ in event 'C'. If these medals went to total $60$ men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?