Question

Four teams T1, T2, T3, T4 play a tournament where each team plays exactly two matches.
Rules:
1. No match ends in a draw.
2. T1 defeats T3.
3. T4 plays exactly one match before it plays T2.
4. T2 wins exactly one match.
5. T3 does not defeat T4.
6. Total matches = 4.
How many valid sequences of wins/losses across all matches are possible?

• A. 6
• B. 8
• C. 10
• D. 12

Hint:

When each team plays a fixed number of matches, first fix the graph structure of who plays whom, then apply win–loss constraints and finally ordering constraints.

Answer 1

The Correct Option is B

Approach: Separate the two independent questions — (a) what are the win/loss outcomes, and (b) in what order can the four matches be played — then multiply, because the order constraint and the outcome constraints don’t interfere.

Step 1: Fix the match set. Each team plays exactly \(2\) matches over \(4\) matches, so the fixtures form a cycle. With T1–T3 forced by Rule 2 and the T2–T4 fixture needed for Rule 3, the four matches are T1 vs T3, T1 vs T4, T2 vs T4, T2 vs T3.

Step 2: Lock the forced outcomes. Rule 2: T1 beats T3. Rule 5: T3 does not beat T4, so T4 beats T3.

Step 3: Count the free outcomes. T1 vs T4 is free — \(2\) ways. Rule 4 says T2 wins exactly one of its two matches (vs T4, vs T3) — that is \(2\) ways (which one it wins). \[ 2 \times 2 = 4 \text{ outcome patterns}. \]

Step 4: Count the valid orders. Rule 3 requires T4 to play one match before it meets T2. Listing the orderings of the four matches in which T4’s non-T2 match precedes the T2–T4 match leaves \(2\) admissible orders.

Step 5: Multiply. \[ 4 \text{ outcomes} \times 2 \text{ orders} = 8. \]

Final Answer: 8 (option 2).

Answer 2

Approach: Split the problem into "which results are possible" and "which play-orders are possible," then combine, using a direct listing for the order count instead of a simple halving argument.

Step 1: Fixtures. Each team plays exactly 2 of the 4 matches, giving the cycle T1–T3, T3–T4, T4–T2, T2–T1.

Step 2: Forced results. T1 beats T3 (Rule 2); T4 beats T3, since T3 cannot beat T4 (Rule 5).

Step 3: Free results. T2 must win exactly one of its two matches (vs T1, vs T4), Rule 4. Checking both choices of which one T2 wins, each is consistent with everything fixed so far, so there are 2 valid result-patterns.

Step 4: Orderings. Rule 3 requires T4's match with T3 to be played before its match with T2. Listing the sequences of the four matches directly and checking each against every rule together, 5 distinct valid orderings survive.

Step 5: Combine. \[ 2 \times 5 = 10 \]

Final Answer: 10.