Question

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:

	TEAM 1	TEAM 2
MATCH 1	P and X	Q and R
MATCH 2	P and R	X and Y
MATCH 3	R and X	Q and Y

Which one of the following options is correct?

• A. P does not have any parent playing
• B. Q does not have any parent playing
• C. R does not have any parent playing
• D. X does not have a child playing

Hint:

Use the fact that a parent's team cannot include his or her own child to test each match one at a time.

Answer 1

The Correct Option is C

Step 1: Set up the parent-child assignment.
X and Y are each the parent of exactly one of P, Q, R, and these two children are different from each other (the child of X is not the child of Y). So two of the three children have a parent in the tournament, and exactly one child has no parent playing. Call the child of X as \(C_X\), the child of Y as \(C_Y\), and the child with no parent as \(C_0\).

Step 2: Use Match 1 to rule out one possibility.
In Match 1, Team 1 is "P and X". Rule (i) says a parent cannot be on the same team as his or her own child. If \(C_X\) were P, then X and his child P would both be on Team 1, breaking rule (i). So \(C_X\) cannot be P; it must be Q or R.

Step 3: Use Match 2 to narrow down \(C_0\).
In Match 2, all four of P, R, X, and Y are playing (Q sits out). If both \(C_X\) and \(C_Y\) were chosen from \(\{P,R\}\), then both parent-child pairs would be present together in this one match, since both children and both parents are all playing at once. That directly breaks rule (ii), which allows at most one parent-child pair per match. So \(C_X\) and \(C_Y\) cannot both lie in \(\{P,R\}\), which means \(C_0\) cannot be Q; the child left without a parent must be P or R.

Step 4: Combine Steps 2 and 3.
From Step 2, \(C_X \in \{Q,R\}\). From Step 3, \(C_0 \in \{P,R\}\), meaning Q is definitely one of \(C_X\) or \(C_Y\) (Q always has a parent playing).
This leaves two possible cases:
Case 1: \(C_0=P\), so \(\{C_X,C_Y\}=\{Q,R\}\).
Case 2: \(C_0=R\), so \(\{C_X,C_Y\}=\{P,Q\}\), and since \(C_X\neq P\) (Step 2), we must have \(C_X=Q\) and \(C_Y=P\).

Step 5: Test Case 1 against Match 3.
In Match 3, Team 1 is "R and X" and Team 2 is "Q and Y" (P sits out). In Case 1, \(C_X\) is Q or R.
If \(C_X=R\), then X and his child R are on the same Team 1, breaking rule (i). So within Case 1 we would need \(C_X=Q\) and \(C_Y=R\).
But then check rule (ii) in Match 3: X's child Q is present (Team 2) along with X (Team 1), giving one parent-child pair; and Y's child R is present (Team 1) along with Y (Team 2), giving a second parent-child pair. Two pairs appear together in the same match, which breaks rule (ii).
So Case 1 fails no matter how we assign Q and R between \(C_X\) and \(C_Y\).

Step 6: Confirm Case 2 works.
In Case 2, \(C_X=Q\), \(C_Y=P\), and \(C_0=R\) (R has no parent playing). Check all three matches:
Match 1 (P,X vs Q,R): X's child Q is on the opposite team from X, so rule (i) holds. Y is not playing, so there is nothing to check for Y. Only one parent-child pair (X-Q) appears, satisfying rule (ii).
Match 2 (P,R vs X,Y): the two children P, R are together on one team and the two parents X, Y are together on the other, so rule (i) is automatically satisfied. Q is sitting out, so only Y's child P is present alongside Y, giving exactly one parent-child pair, satisfying rule (ii).
Match 3 (R,X vs Q,Y): X's child Q is on the opposite team from X, satisfying rule (i). P is sitting out, so Y's child P is not present in this match, leaving only one parent-child pair (X-Q), satisfying rule (ii).
Every rule holds in every match, so Case 2 is the valid arrangement.

Step 7: Read off the answer.
In this valid arrangement, R is the child with no parent playing, X's child is Q, and Y's child is P.
Checking the options: (A) is false since P has a parent (Y). (B) is false since Q has a parent (X). (D) is false since X's child Q does play in Matches 1 and 3. Only (C) is true.

Step 8: Final conclusion.
\[ \boxed{\text{R does not have any parent playing}} \]
Hence the correct option is (C).