Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ items from a set of $n$ items (without regard to order) is given by ${}^nC_r = \frac{n!}{r!(n-r)!}$.
• Problem Interpretation: Understanding the constraints of the problem, especially "no couple played in a match," is crucial for setting up the correct equation.

Step-by-Step Solution

Step 1: Define the variable Let $n$ be the number of couples who participated in the tournament. Thus, there are $2n$ people in total.

Step 2: Formulate the equation representing the number of matches We need to select two men and two women such that no couple plays in a match. First, we select 2 couples out of $n$ in ${}^nC_2$ ways. Then from each couple, we select one person. Number of ways of doing this is $2 \times 2 = 4$. However, this also includes cases where a couple plays each other. To exclude a couple playing against each other, we need to select 2 couples out of n couples, then select one person from each couple such that we don't have a couple playing.

The number of ways to choose 2 men from $n$ men and 2 women from $n$ women such that no couple plays against each other is calculated as follows: Select 2 couples from $n$ couples: ${}^nC_2$ From the first couple, we can choose the man or woman (2 choices). From the second couple, we can choose the man or woman (2 choices). So, we have $2 \times 2 = 4$ choices. Thus, the number of ways to choose the players is ${}^nC_2 \times 2 \times 2 = 4 \times {}^nC_2$. This gives us the number of matches. The problem states that the number of matches is 840. However, this is not correct. Instead, we choose 2 men out of $n$ men and 2 women out of $n$ women. The number of ways of selecting 2 men out of $n$ is ${}^nC_2$. After selecting two men, we select 2 women out of the remaining $n-2$ women, which is ${}^{n-2}C_2$. The number of ways of doing this is ${}^nC_2 \times {}^{n-2}C_2 \times 2 = 840/2 = 420$. The total number of matches is given by $2 \times {}^nC_2 \times {}^{n-2}C_2$. Thus we have ${}^nC_2 \times {}^{n-2}C_2 = 420$.

Step 3: Simplify the equation ${}^nC_2 \times {}^{n-2}C_2 = 420$ $\frac{n!}{2!(n-2)!} \times \frac{(n-2)!}{2!(n-4)!} = 420$ $\frac{n(n-1)}{2} \times \frac{(n-2)(n-3)}{2} = 420$ $n(n-1)(n-2)(n-3) = 4 \times 420 = 1680$

Step 4: Solve for n We are looking for an integer solution for $n$. We can try integer values of $n$. If $n=5$, then $5 \times 4 \times 3 \times 2 = 120$. Too small. If $n=6$, then $6 \times 5 \times 4 \times 3 = 360$. Too small. If $n=7$, then $7 \times 6 \times 5 \times 4 = 840$. Too small. If $n=8$, then $8 \times 7 \times 6 \times 5 = 1680$. This is the solution. Therefore, $n=8$.

Step 5: Calculate the total number of persons Since $n$ is the number of couples, the total number of persons is $2n = 2 \times 8 = 16$.

Common Mistakes & Tips

• Incorrect Formula: Double-check the formulas for combinations and permutations.
• Misinterpreting the Constraints: The condition "no couple played in a match" is crucial. Make sure your formula accounts for this.
• Trial and Error: When solving equations like $n(n-1)(n-2)(n-3) = 1680$, trial and error with nearby integer values of $n$ is often the fastest approach.

Summary We defined $n$ as the number of couples. Then, we set up the equation ${}^nC_2 \times {}^(n-2)C_2 = 420$ representing the number of ways to form matches with the given constraint. Solving for $n$, we found $n=8$. The total number of persons is $2n = 16$.

Final Answer The final answer is \boxed{16}.