Question

If a standard knockout tournament is being organized for a total of 19 teams, what is the total number of matches that will be played in the tournament, and how many teams will receive a 'Bye' in the first round?

• A. \( 18 \text{ matches and } 13 \text{ byes} \)
• B. \( 19 \text{ matches and } 13 \text{ byes} \)
• C. \( 18 \text{ matches and } 3 \text{ byes} \)
• D. \( 20 \text{ matches and } 5 \text{ byes} \)

Hint:

To save time during competitive shifts, remember that the total matches in *any* standard single-elimination knockout tournament is simply \( N - 1 \). This immediately narrows your options down without extra calculations.

Answer 1

The Correct Option is A

Concept: Tournament planning requires specific structural formulas to arrange brackets efficiently. For a knockout tournament:

• Total matches played (\( M \)) is always one less than the number of participating teams (\( N \)): \[ M = N - 1 \]
• Total byes distributed is calculated by subtracting the number of teams from the next highest power of two (\( 2^n \)): \[ \text{Total Byes} = 2^n - N \]

Step 1: Calculate the total number of matches.
Given the number of competing teams is \( N = 19 \): \[ \text{Matches} = N - 1 = 19 - 1 = 18 \text{ matches} \]

Step 2: Determine the number of byes using powers of two.
The powers of two are \( 2, 4, 8, 16, 32, \dots \) The next power of two strictly greater than 19 is 32 (\( 2^5 \)). Subtracting our team count from this boundary: \[ \text{Total Byes} = 32 - 19 = 13 \text{ byes} \]