IPMAT Indore 2023 · Quantitative Ability · Question 19

Step 1 — check the total. With 5 players each meeting every other once, the number of games is

\left( \genfrac{}{}{0pt}{}{5}{2}\right)=10

Each game awards exactly one point, so every valid sequence must sum to 10. All four options do, so the total is not the discriminator — something sharper is needed.

Step 2 — look at option (d): 4, 4, 1, 1, 0. A player scoring 4 has won all four of their games — including the game against the other player who scored 4.

But that second player also scored 4, so they too won all four of their games, including the one against the first player.

They cannot both have won the game they played against each other. Contradiction.

Step 3 — confirm the others are achievable.

(a) 2,2,2,2,2 — a cyclic arrangement: each player beats the next two and loses to the previous two.

(b) 3,3,2,1,1 and (c) 3,2,2,2,1 — both are realisable; an exhaustive check over all {2}^{10}=1024 outcome patterns confirms each occurs.

Answer: Option 4 — 4, 4, 1, 1, 0

Shortcut. The maximum score is 4, and at most one player can achieve it. Two players on the maximum is an instant contradiction, because the maximum means “beat everyone”, and they played each other.

That single observation answers the question without checking totals or constructing tournaments.

Generalise it: in an n-player round robin with no draws, at most one player can score n-1, and at most one can score 0. Any sequence listing the top score twice — or the bottom score twice — is impossible. Scan the extremes first; the sum-to-\left( \genfrac{}{}{0pt}{}{n}{2}\right) check rarely separates the options, because setters make sure all four sum correctly.