DAMATH is a board game that incorporates mathematical skills in the Filipino game Dama. In a school DAMATH tournament, there are 28 participants who are divided into 7 groups. Each participant plays against each member of his group in the eliminations. The winner in each group advances to the semi-finals where they again compete with each other. The five players with the most number of wins proceed to the final round and play against each other. Assume that there are no ties.
What is the total number of games to be played in the eliminations?
How many matches will be played in the final round?
In how many possible ways can the top five players in the semifinals come up?
In how many possible ways can the 1st, 2nd, and 3rd placer be declared in the final round?
How many matches will be played altogether?
The Correct Answer and Explanation is:
1. Total number of games in the eliminations
There are 7 groups, and 28 participants are evenly divided.
So, each group has: 287=4 players\frac{28}{7} = 4 \text{ players}
In a group of 4 players, each player plays against every other player once.
This is a combination problem: (42)=6 games per group\binom{4}{2} = 6 \text{ games per group}
For 7 groups: 7×6=42 games7 \times 6 = \boxed{42 \text{ games}}
2. Matches in the final round
Top 5 players compete against each other, so again use: (52)=10 games\binom{5}{2} = \boxed{10 \text{ games}}
3. Possible ways to select top five players in the semifinals
Only 7 players make it to the semifinals (1 per group). From these, we choose 5: (75)=21 ways\binom{7}{5} = \boxed{21 \text{ ways}}
4. Ways to declare 1st, 2nd, and 3rd place in final round
This is a permutation of 3 places out of 5: P(5,3)=5×4×3=60 waysP(5, 3) = 5 \times 4 \times 3 = \boxed{60 \text{ ways}}
5. Total matches played altogether
Breakdown:
- Eliminations: 42 games
- Semifinals: 7 players →
(72)=21 games\binom{7}{2} = 21 \text{ games}
- Finals: 5 players →
(52)=10 games\binom{5}{2} = 10 \text{ games}
Total: 42+21+10=73 games42 + 21 + 10 = \boxed{73 \text{ games}}
The DAMATH tournament follows a structured format involving elimination, semi-final, and final rounds. Understanding the total number of games played requires applying concepts from combinatorics, particularly combinations and permutations.
In the elimination round, 28 players are divided into 7 groups of 4. Each participant in a group plays every other member once. The number of matches in each group is the combination of 4 players taken 2 at a time, or (42)=6\binom{4}{2} = 6. Across all 7 groups, the total becomes 7×6=427 \times 6 = 42 games.
From each group, the winner proceeds to the semi-finals, giving 7 players. In this round, each player again plays every other once, totaling (72)=21\binom{7}{2} = 21 games.
From these 7 semifinalists, the top 5 in terms of wins move to the finals. The number of ways to choose 5 players from 7 is (75)=21\binom{7}{5} = 21 ways. Once in the finals, the 5 players face each other in matches, totaling (52)=10\binom{5}{2} = 10 games.
To determine the ranking (1st, 2nd, 3rd) from the 5 finalists, we calculate the number of permutations of 3 players from 5: P(5,3)=5×4×3=60P(5, 3) = 5 \times 4 \times 3 = 60 ways.
Adding all matches from every stage gives a total of 42+21+10=7342 + 21 + 10 = 73 games. These calculations provide a comprehensive analysis of the tournament’s match structure and outcome possibilities.
