In a tournament of 7 players, each player plays every other player once. How many matches are there?

Question

In a tournament of 7 players, each player plays every other player once. How many matches are there?

1. A. 21 (Correct answer)
2. B. 42
3. C. 36
4. D. 28

Correct Answer

Option A — 21

Detailed Solution & Explanation

The correct answer is 21.

Key Points

• Every match is an unordered pair of players, so the count is C(7, 2).
• C(7, 2) = (7 × 6) / 2 = 21.
• Dividing by 2 matters: 7 × 6 = 42 counts each match twice (A vs B and B vs A), which is exactly option B.
• For a single round-robin among n players the answer is always n(n − 1)/2.
• The same formula answers the common variants: a double round-robin doubles it to n(n − 1), and the number of handshakes in a room of n people is again n(n − 1)/2.
• For 7 players that means 21 single-leg matches, or 42 if every pair meets twice — which is exactly why option B is offered.

Additional Information

• A single round-robin among n players needs n(n − 1)/2 matches; a double round-robin needs n(n − 1).
• The identical formula counts handshakes in a room of n people, diagonals-plus-sides of a polygon, and lines through n points with no three collinear.
• The division by 2 reflects that a pair is unordered — A versus B and B versus A are the same fixture, which is exactly what the doubled distractor tests.

प्रश्न (हिन्दी में)

7 खिलाड़ियों के एक टूर्नामेंट में, प्रत्येक खिलाड़ी दूसरे खिलाड़ी से एक बार खेलता है। कुल कितने मैच हैं?

1. A. 21
2. B. 42
3. C. 36
4. D. 28