Eight teams take part 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 organizers decide to proceed, nonetheless, with the remaining matches. The total number of matches that were played in the tournament that year is
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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.
Tournament Stat
Value
Math Logic
Total Teams
8
Base roster (n=8)
Total Matches (No Bans)
28
8C2 combination formula
Intended Matches per Team
7
n−1 (You play everyone except yourself)
Matches Played by Suspended Team
3
Stated in prompt
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 nC2:
8C2=28×78C2=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:
7C2=27×6=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
In a chess tournament there are 5 contestants. Each player plays against all the others exactly once. No game results in a draw. The winner in a game gets one point and the loser gets zero points. Which of the following sequences cannot represent the scores of the five players?
A2, 2, 2, 2, 2
B3, 2, 2, 2, 1
C3, 3, 2, 1, 1
D4, 4, 1, 1, 0
Pick an option to attempt
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 (25)=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 s1≤s2≤⋯≤sn represents a valid round-robin tournament if and only if:
1. The sum of all scores equals (2n).
2. For any subset of k players (sorted from lowest to highest), their combined score must be at least (2k), representing the matches they played against each other.
i=1∑ksi≥(2k) for all k=1,2,…,nStep 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: s1=0≥(21)=0. (Passes)
* Check k=2: s1+s2=0+1=1≥(22)=1. (Passes)
* Check k=3: s1+s2+s3=0+1+1=2≥(23)=3.
Here, 2≯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
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
Teams
Played
Wins
Losses
Draws
Points
A
5
0
8
B
5
2
6
C
5
2
5
D
5
1
5
E
5
1
F
5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
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Total number of matches ending in draw is
A12
B4
C5
D6
Pick an option to attempt
The Setup: Six teams, every pair meets once, so (26)=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 15×2=30 points - that fixed total is the key that opens the table. For any team, played = wins + losses + draws =5 and points =2(wins)+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 pointsStep 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.
Teams
Played
Wins
Losses
Draws
Points
A
5
3
0
2
8
B
5
3
2
0
6
C
5
2
2
1
5
D
5
1
1
3
5
E
5
0
1
4
4
F
5
0
3
2
2
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:
A
B
C
D
E
F
A
-
W
W
D
D
W
B
L
-
W
L
W
W
C
L
L
-
W
D
W
D
D
W
L
-
D
D
E
D
L
D
D
-
D
F
L
L
L
D
D
-
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:
212=6Final Answer: 6
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
Teams
Played
Wins
Losses
Draws
Points
A
5
0
8
B
5
2
6
C
5
2
5
D
5
1
5
E
5
1
F
5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
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Which team has the highest number of draws
AA
BC
CD
DE
Pick an option to attempt
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.
Teams
Played
Wins
Losses
Draws
Points
A
5
3
0
2
8
B
5
3
2
0
6
C
5
2
2
1
5
D
5
1
1
3
5
E
5
0
1
4
4
F
5
0
3
2
2
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
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
Teams
Played
Wins
Losses
Draws
Points
A
5
0
8
B
5
2
6
C
5
2
5
D
5
1
5
E
5
1
F
5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
show less
Total points Team F scored was
A0
B1
C2
D3
Pick an option to attempt
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.
Teams
Played
Wins
Losses
Draws
Points
A
5
3
0
2
8
B
5
3
2
0
6
C
5
2
2
1
5
D
5
1
1
3
5
E
5
0
1
4
4
F
5
0
3
2
2
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 pointsStep 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
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
Teams
Played
Wins
Losses
Draws
Points
A
5
0
8
B
5
2
6
C
5
2
5
D
5
1
5
E
5
1
F
5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
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Which team was not defeated by team A
AB
BC
CD
DF
Pick an option to attempt
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.
Teams
Played
Wins
Losses
Draws
Points
A
5
3
0
2
8
B
5
3
2
0
6
C
5
2
2
1
5
D
5
1
1
3
5
E
5
0
1
4
4
F
5
0
3
2
2
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
In a football tournament six teams A, B, C, D, E, and F participated. Every pair of teams had exactly one match among them. For any team, a win fetches 2 points, a draw fetches 1 point, and a loss fetches no points. Both teams E and F ended with less than 5 points. At the end of the tournament points table is as follows (some of the entries are not shown):
Teams
Played
Wins
Losses
Draws
Points
A
5
0
8
B
5
2
6
C
5
2
5
D
5
1
5
E
5
1
F
5
It is known that: (1) team B defeated team C, and (2) team C defeated team D
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Team E was defeated by
ATeams A and B only
BOnly Team A
COnly Team B
DTeams A, B and D only
Pick an option to attempt
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.
Teams
Played
Wins
Losses
Draws
Points
A
5
3
0
2
8
B
5
3
2
0
6
C
5
2
2
1
5
D
5
1
1
3
5
E
5
0
1
4
4
F
5
0
3
2
2
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