IPMAT Indore Tournaments — PYPs

The Setup:
We are dealing with a round-robin tournament combinatorics problem. The base logic requires calculating the total number of matches in a perfect bracket, and then subtracting the matches that got canceled because a team got hit with the ban hammer. I have set up a stat box to extract the core data before we run the calculations.

Step 1: Extracting the Meta (Stat Box)
Let's map out exactly what is happening in this bracket before the suspension ruins the schedule.

Step 2: The Base Timeline
In a flawless, perfectly run round-robin tournament, every team plays every other team exactly once. The formula to find total unique handshakes (or matches) is $^nC_2$:
$$^8C_2 = \frac{8 \times 7}{2}$$
$$^8C_2 = 28$$
So, the organizers originally scheduled exactly 28 matches.

Step 3: Calculating the Dropped Matches
One team gets toxic and gets suspended.
Every single team in this tournament was scheduled to play 7 matches.
This suspended team managed to play exactly 3 matches before getting kicked.
To find the number of canceled matches, subtract what they played from what they were supposed to play:
$$7 - 3 = 4$$
Exactly 4 scheduled matches literally evaporated from the bracket.

Step 4: The Final Tally
Subtract the canceled matches from the total base schedule to find out how many matches actually happened.
$$28 - 4 = 24$$
There were 24 matches officially played that year.

Step 5: The Audit (Double Check Protocol)
Running it back using an alternative method to guarantee we aren't throwing.
Let's look at the tournament from the perspective of the 7 teams that *didn't* get banned.
If those 7 teams just played a private tournament among themselves, the total matches would be:
$$^7C_2 = \frac{7 \times 6}{2} = 21$$
Now, we just add the matches that the banned team played against 3 of these valid teams before getting kicked:
$$21 + 3 = 24$$
Both methods independently output exactly 24. The logic is completely locked in and validated. No cap.

Final Answer: 24

The Setup: A round-robin chess tournament features $5$ players where every player plays everyone else exactly once. A win awards $1$ point, a loss awards $0$ points, and no draws occur. We must identify which sequence of scores is mathematically impossible.

Step 1: Establish the total points in the system.
The number of players is $n = 5$. The total number of matches played in the tournament is $\binom{5}{2} = 10$.
Since each match awards exactly $1$ point to the winner and $0$ to the loser, exactly $10$ points are distributed across the 5 players. Every valid score sequence must sum to $10$.

Step 2: Apply Landau's Theorem for score sequences.
Landau's Theorem states that a sequence of $n$ scores $s_1 \le s_2 \le \dots \le s_n$ represents a valid round-robin tournament if and only if:
1. The sum of all scores equals $\binom{n}{2}$.
2. For any subset of $k$ players (sorted from lowest to highest), their combined score must be at least $\binom{k}{2}$, representing the matches they played against each other.

$$\sum_{i=1}^{k} s_i \ge \binom{k}{2} \text{ for all } k=1, 2, \dots, n$$

Step 3: Evaluate the given options against Landau's criteria.
Let's test the suspected invalid sequence $4, 4, 1, 1, 0$.
Sort the sequence in ascending order: $s = (0, 1, 1, 4, 4)$.
* Check $k=1$: $s_1 = 0 \ge \binom{1}{2} = 0$. (Passes)
* Check $k=2$: $s_1 + s_2 = 0 + 1 = 1 \ge \binom{2}{2} = 1$. (Passes)
* Check $k=3$: $s_1 + s_2 + s_3 = 0 + 1 + 1 = 2 \ge \binom{3}{2} = 3$.
Here, $2 \ngtr 3$. The lowest three players must have generated at least $3$ points purely from playing against each other, making a combined score of $2$ physically impossible.

Final Answer: 4, 4, 1, 1, 0

The Setup: Six teams, every pair meets once, so $\binom{6}{2}=15$ matches. Each match hands out exactly 2 points (2-0 for a decision, 1-1 for a draw), so the whole tournament is worth $15 \times 2 = 30$ points - that fixed total is the key that opens the table. For any team, played $=$ wins $+$ losses $+$ draws $=5$ and points $=2(\text{wins}) + \text{draws}$.

Step 1: Decode A through D. Two equations per team, and both are already given (losses and points):
* A - $0$ losses, so $w+d=5$; with $2w+d=8$, subtracting gives $w=3, d=2$.
* B - $2$ losses, so $w+d=3$; with $2w+d=6$, $w=3, d=0$.
* C - $2$ losses, so $w+d=3$; with $2w+d=5$, $w=2, d=1$.
* D - $1$ loss, so $w+d=4$; with $2w+d=5$, $w=1, d=3$.

Step 2: E is forced by the under-5 clause. E has $1$ loss, so $w+d=4$ and points $=2w+d=w+4$. For E to stay under 5 points we need $w<1$, so $w=0$, $d=4$, and E finishes on 4 points - a single win anywhere would break the clause.

Step 3: F falls out of the 30-point total.

$$F = 30 - (8+6+5+5+4) = 30 - 28 = 2 \text{ points}$$

Step 4: Split F's 2 points - and this is the step worth slowing down on. Two ways to score 2: $1$ win, $0$ draws, $4$ losses; or $0$ wins, $2$ draws, $3$ losses. Look at E to kill the first one. E must draw *four* of its five matches. B has $0$ draws, so B is not one of E's partners. If F also had $0$ draws, E's only available partners would be A, C and D - three teams for four draws, which is impossible. So F must carry draws, and **F is $0$ wins, $2$ draws, $3$ losses.**

Step 5: The completed table.

Step 6: The table plus the two given results actually pins every single match. E draws A, C, D and F and loses to B (Step 4). D's one loss is to C, so its three draws are A, E, F and it beats B. A never loses and draws only D and E, so it beats B, C and F. C's two losses are to A and B; its win over D and its draw with E are known, so it beats F. That leaves F losing to A, B and C. Master Matrix - each cell is the row team's result:

Step 7: Count the draws. The Draws column totals $2+0+1+3+4+2=12$, but every drawn match is counted twice - once for each team. So the number of matches that ended in a draw is:

$$\frac{12}{2}=6$$

Final Answer: 6

The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.

Step 1: Read straight down the Draws column: A has $2$, B has $0$, C has $1$, D has $3$, E has $4$, F has $2$.

Step 2: E tops it with $4$ draws, and it is not close - D is next on $3$. This is no accident: E is the team squeezed by the under-5-points clause, which forced four of its five matches to be draws.

Final Answer: E

The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.

Step 1: The tournament is worth a fixed $30$ points - $15$ matches, each handing out exactly $2$ points whether it is decided or drawn.

Step 2: Subtract everyone else. A, B, C, D and E take $8+6+5+5+4=28$ between them, so F is left with:

$$30-28=2 \text{ points}$$

Step 3: Sanity-check it against F's row: $0$ wins and $2$ draws is $2(0)+2=2$ points. Consistent, and comfortably under the 5-point cap the question imposes on F.

Final Answer: 2

The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.

Step 1: A won $3$ matches, so exactly two teams escaped it - and the options only offer B, C, D and F, so we need to find which of those four A did not beat.

Step 2: Go to D's row: D has exactly one loss. The question stem already tells us *who* dealt it - team C defeated D. That single loss is therefore spoken for.

Step 3: So D cannot have lost to A as well. A's match against D was one of A's two draws, which means D is a team A did not defeat.

Step 4: Check the others quickly. B and C both sit on $2$ losses; A is unbeaten and B's and C's losses are accounted for by A and each other (B beat C, so C's other loss is to A, and B's two losses are to A and D). F lost $3$, including to A. D is the only option left standing. (E also escaped A, via a draw, but E is not on the menu.)

Final Answer: D

The Setup: This is the same tournament table as the rest of the set. The full derivation lives on the first question of the set; here is the finished table it produces, which is all we need.

Step 1: E's row shows exactly one loss. So E was beaten by exactly one team - that alone wipes out 'Teams A and B only' and 'Teams A, B and D only', which both name two or more.

Step 2: Now find the one. E drew its other four matches. B is the only team in the whole table with zero draws, so B cannot be one of E's four draw partners - the only remaining relationship between B and E is a result, and E does not win at all ($0$ wins).

Step 3: Therefore B beat E, and since E has just the one loss, no one else did. A drew with E rather than beating it, which is exactly why 'Only Team A' fails.

Final Answer: Only Team B