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: List the three possible cases for who has no parent playing.
Exactly one of P, Q, R has no parent in the tournament, so there are three cases to test: no-parent child is P, or Q, or R. In each case, the other two children are shared between X and Y as their respective children, in some order.

Step 2: Test "Q has no parent."
Then P and R are the children of X and Y, in some order. Look at Match 2, where P, R, X, and Y all play together (P, R on one team, X, Y on the other). Since both P and R are children of the two parents present, both parent-child pairs would appear together in this single match, which is not allowed since a match may contain at most one such pair. This case fails.

Step 3: Test "P has no parent."
Then Q and R are the children of X and Y, in some order. Match 1 has Team 1 as "P and X" and Team 2 as "Q and R". If X's child were R, look at Match 3 (Team 1: R, X; Team 2: Q, Y): X would share a team with his own child R, breaking the same-team rule. So X's child must be Q here, leaving R as Y's child. Now recheck Match 3: X's child Q sits on Team 2 while Y's child R sits on Team 1, so BOTH parent-child pairs are present in Match 3 together (X-Q and Y-R), again breaking the one-pair-per-match rule. So this case also fails.

Step 4: Test "R has no parent."
Then P and Q are the children of X and Y, in some order. From Match 1 (Team 1: P, X), if X's child were P they would share a team, which is forbidden, so X's child must be Q, leaving P as Y's child.
Check Match 2 (Team 1: P, R; Team 2: X, Y): Q is sitting out, so X's child Q is absent from this match, and only Y's child P is present alongside Y. That is one pair, which is allowed.
Check Match 3 (Team 1: R, X; Team 2: Q, Y): P is sitting out, so Y's child P is absent from this match, and only X's child Q is present alongside X, on the opposite team. That is again one pair, allowed, and no same-team conflict.
Every match checks out cleanly in this case.

Step 5: Conclude.
Since the first two cases both broke a rule and this last case satisfies every match, R must be the child without a parent playing, with X's child being Q and Y's child being P.
\[ \boxed{\text{R does not have any parent playing}} \]