IPMAT Indore 2023 (SA) - In a chess tournament, there are four groups, each containing an equal number of players. Each player plays 1. against every other player belonging to one's own group exactly once; 2. against each player belonging to one of the remaining three groups exactly twice; 3. against each player belonging to one of the remaining two groups exactly three times; and 4. against each player belonging to the remaining group exactly four times. If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is | PYQs + Solutions

IPMAT Indore 2023

Modern Math

Permutation & Combination

Medium

In a chess tournament, there are four groups, each containing an equal number of players. Each player plays
1. against every other player belonging to one's own group exactly once; $\newline$ 2. against each player belonging to one of the remaining three groups exactly twice; $\newline$ 3. against each player belonging to one of the remaining two groups exactly three times; and $\newline$ 4. against each player belonging to the remaining group exactly four times.
If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is

Entered answer:

✅ Correct Answer: 4

There are four groups, and each player plays against players from other groups as per the following rules:

1. Matches within the group:

Each player plays against every other player in their own group exactly once. For a group with

n

players, the number of matches within the group is:

{}^{n}C_2 = \frac{n(n-1)}{2}

Since there are four groups, the total number of matches within all groups is:

4 \times \frac{n(n-1)}{2} = 2n(n-1)

2. Matches with players in the other three groups:

Each player plays against each player in the remaining three groups exactly twice. The total number of players in the other three groups is

3n

. Therefore, each player plays

3n

players twice, resulting in:

2 \times 3n = 6n

matches per player

Since there are

4n

players in total, the total number of matches between all players and players from other groups is:

4n \times 6n = 24n^2

3. Matches with players in two other groups: Each player plays against each player in two of the remaining groups exactly three times. The total number of players in two other groups is

2n

, and the number of matches per player with them is:

3 \times 2n = 6n

matches per player

Again, for all players, the total number of such matches is:

4n \times 6n = 24n^2

4. Matches with players in the last group: Each player plays against each player in the remaining group exactly four times. The number of players in this group is

n

, and each player plays

4n

matches against them:

4 \times n = 4n

matches per player

Therefore, the total number of matches between all players and players from the last group is:

4n \times 4n = 16n^2

Now, summing all the matches:

\text{Total matches} = 2n(n-1) + 24n^2 + 24n^2 + 16n^2 = 66n^2 - 2n

We want the total number of matches to be greater than 1000, so:

66n^2 - 2n > 1000

Solving this inequality results in

n = 4

Thus, the minimum number of players in each group is:

\boxed{4}

IPMAT Indore 2023 (SA) PYQs

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