Q.The number of matches in a league (round-robin) tournament of 10 teams, where every team plays every other team once, is:

Given: N = 10 teams, league (round-robin), each team plays every other team once.

Concept understanding — why (N(N-1))/2

In a league (round-robin) tournament every team meets every other team. Count it in two steps:

1. Take any one team. It has to play the other N - 1 teams. With 10 teams, each team plays 9 matches.
2. There are N teams, so that is N × (N-1) = 10 × 9 = 90 team-fixtures.
3. But each match involves two teams, so every match has been counted twice (once from A's side, once from B's side). Divide by 2.

Total matches = (N(N-1))/2 = (10 × 9)/2 = 45

Step-by-step

1. N = 10
2. N - 1 = 9 (each team's number of matches)
3. N × (N-1) = 10 × 9 = 90 (every match counted twice)
4. 90/2 = 45 matches …