IPMAT Indore 2025 · Quantitative Ability · Question 10

IPMAT Indore 2025 previous year question on Permutation and Combination, with the answer and a step-by-step solution.

Permutation and CombinationModern MathsType in the answer (TITA)Easy
Eight teams participate in a tournament where each team plays against every other team exactly once. In a particular year, one team got suspended after playing 3 matches, due to a disciplinary issue. The organisers decided to proceed, nonetheless, with the remaining matches. The total number of matches that were played in the tournament that year is
Answer & solution

Answer: 24

Step 1 — the full schedule. Every pair of teams meets once, so the number of matches is the number of ways of choosing 2 teams out of 8:

\left( \genfrac{}{}{0pt}{}{8}{2}\right)=\frac{8\times 7}{2}=28

Step 2 — what the suspension costs. The suspended team was supposed to play 8-1=7 matches. It actually played 3, so

7-3=4\text{matches were never played.}

Every other match in the schedule involves two surviving teams, so all of those still went ahead.

Step 3 — subtract.

28-4=24

Answer: 24

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