Three children P, Q, R and two grown-ups X, Y play a badminton doubles tournament. X and Y are parents to two of the children playing. The child of X is not the same as the child of Y. Exactly one of the children does not have a parent playing in the tournament. The following rules are followed:

(i) A parent and his/her child cannot be on the same team.

(ii) A match can feature at most one parent and his/her child, that is, a maximum of one parent-child pair can play in a match.

The following matches were played:

Which one of the following options is correct?

	TEAM 1	TEAM 2
$\text{MATCH 1}$	$P\text{and}X$	$Q\text{and}R$
$\text{MATCH 2}$	$P\text{and}R$	$X\text{and}Y$
$\text{MATCH 3}$	$R\text{and}X$	$Q\text{and}Y$

The correct option is: R does not have any parent playing

Let's analyze the given rules and the matches played to determine the parent-child relationships step-by-step.

1. Understand the Setup and Constraints:
- There are three children: $P$, $Q$, $R$.
- There are two adults: $X$, $Y$.
- $X$ and $Y$ are parents to two distinct children. Let's denote $C\left(X\right)$ as the child of $X$ and $C\left(Y\right)$ as the child of $Y$.
- We are given: $C\left(X\right)\ne C\left(Y\right)$.
- Exactly one of the three children does not have a parent playing in the tournament. This means two children have a parent (one has $X$ as a parent, and another has $Y$ as a parent).

Rules:
- Rule (i): A parent and child cannot be on the same team.
- Rule (ii): A match can feature at most one parent-child pair playing (either on opposing teams, since they cannot be on the same team).

2. Analyze the Matches:

Match 1: Team 1 ($P$ and $X$) vs Team 2 ($Q$ and $R$)
- By Rule (i), since $P$ and $X$ are on the same team, $P$ cannot be the child of $X$. Therefore:
$C\left(X\right)\ne P$

Match 2: Team 1 ($P$ and $R$) vs Team 2 ($X$ and $Y$)
- The players participating in this match are $P,R,X,Y$.
- The potential parent-child relationships among the playing participants are $(X,P)$, $(X,R)$, $(Y,P)$, and $(Y,R)$.
- From Match 1, we already know $C\left(X\right)\ne P$. Therefore, the pair $(X,P)$ is not a parent-child relationship.
- Rule (ii) states that a match can feature at most one parent-child pair. This means among the active players in Match 2, at most one parent-child pair is present.

Match 3: Team 1 ($R$ and $X$) vs Team 2 ($Q$ and $Y$)
- By Rule (i), since $R$ and $X$ are partners on Team 1, $R$ cannot be the child of $X$. Therefore:
$C\left(X\right)\ne R$
- By Rule (i), since $Q$ and $Y$ are partners on Team 2, $Q$ cannot be the child of $Y$. Therefore:
$C\left(Y\right)\ne Q$

3. Determine the Parent-Child Relationships:
- Since the child of $X$ must be one of $P,Q,R$, and we found:
$C\left(X\right)\ne P$ and $C\left(X\right)\ne R$
the only remaining choice for $X$'s child is $Q$:
$C\left(X\right)=Q$

Now let's find the child of $Y$:
- The child of $Y$ must be different from the child of $X$, so $C\left(Y\right)\ne Q$ (which is also consistent with Match 3).
- Thus, the child of $Y$ must be either $P$ or $R$.

Let's check the constraints in Match 2 where the active players are $P,R,X,Y$:
- Since $C\left(X\right)=Q$ and $Q$ is not playing in Match 2, $X$'s child is not playing in Match 2.
- If $C\left(Y\right)=P$, then the parent-child pair $(Y,P)$ is playing in Match 2. This is valid as it features exactly one parent-child pair, satisfying Rule (ii).
- If $C\left(Y\right)=R$, then the parent-child pair $(Y,R)$ plays in Match 2, which also satisfies Rule (ii).

Let's check the constraints in Match 3 where the active players are $R,X,Q,Y$:
- We already know $C\left(X\right)=Q$, which means the parent-child pair $(X,Q)$ is playing in this match. This counts as 1 parent-child pair.
- By Rule (ii), a match can feature at most one parent-child pair. Since $(X,Q)$ is already playing, no other parent-child relationship can be active in Match 3.
- The other child playing in Match 3 is $R$, and the other parent is $Y$. Since there can be no other parent-child pair, $R$ cannot be the child of $Y$:
$C\left(Y\right)\ne R$

4. Conclusion:
- Since $C\left(Y\right)\ne R$ and we already know $C\left(X\right)=Q\ne R$, the child $R$ is not the child of either $X$ or $Y$.
- Since $X$ and $Y$ are the only parents playing in the tournament, R does not have any parent playing (making $P$ the child of $Y$).