Some couples participated in a mixed doubles badminton tournament. If the number of matches played,

Q: Wait, why do we multiply by 2 in the final equation? Shouldn't it just be the number of ways to choose the players?

Great observation! Selecting the 4 players is only half the battle. Once you have 2 men ($M_1, M_2$) and 2 women ($W_3, W_4$), you still need to form the teams. You can have ($M_1, W_3$) vs ($M_2, W_4$) OR ($M_1, W_4$) vs ($M_2, W_3$). That is 2 distinct ways to arrange the match for every single group of 4 players. We must account for this!

Q: Why do we select women from $n-2$ couples? Why not just $n$?

Remember the golden rule: no couple can play in the same match. If we have already selected 2 men from 2 different couples, their wives are now 'banned' from the match. If we picked women from the original $n$ couples, we might accidentally pick one of those wives. By removing those 2 couples entirely, we guarantee the 'no couple' constraint is satisfied.

Q: The equation $n(n-1)(n-2)(n-3) = 1680$ looks like a nightmare to solve. Do I really have to expand it into a quartic polynomial?

Absolutely not! That is the trap. In JEE, we look for elegance. Notice that $n(n-1)(n-2)(n-3)$ is the product of 4 consecutive integers. Instead of expanding, factorize 1680. You will find it is $8 \times 7 \times 6 \times 5$. Comparing the terms, $n$ must be 8. Always look for factors before expanding!

Question

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ________.

Sigma Percentile · Accepted Answer

### The Dance of the Couples: A Combinatorial Journey

Welcome, future engineers! Today, we are not just solving a math problem; we are choreographing a badminton tournament. Imagine standing on the sidelines of a mixed doubles tournament where a very specific, mischievous rule is in place: no couple can play in the same match.

This isn't just a rule; it is a beautiful combinatorial constraint that will guide our entire journey.

### Phase 1: Visualizing the Constraint

Let us define our universe. We have $n$ couples, which means we have $2n$ people in total. A mixed doubles match requires 4 people: 2 men and 2 women.

The constraint is absolute: no couple can play in the same match. This means if a husband is on the court, his wife cannot be his partner, nor can she be his opponent.

Physically, this implies that the 4 players on the court must be drawn from 4 completely different couples. If we were to pick players from only 3 couples, at least one couple would be represented by two people, violating our rule. This realization is our first victory.

### Phase 2: The Selection Dance

Now, let us build our match. First, we choose 2 men from our $n$ couples. The number of ways to do this is simply $\binom{n}{2}$. Let us call these men $M_1$ and $M_2$, hailing from couples $C_1$ and $C_2$.

Next, we need 2 women. Because of the 'no couple' rule, the wives of $M_1$ and $M_2$ are strictly forbidden from this match. We must exclude couples $C_1$ and $C_2$ from our selection pool.

This leaves us with $n-2$ couples. From these remaining couples, we select our 2 women. The number of ways to do this is $\binom{n-2}{2}$.

### Phase 3: The Arrangement Factor

We have our 4 players: $M_1, M_2, W_3, W_4$. Are we done? Not quite. We have the players, but we need to form the teams.

For any fixed set of 4 players, there are 2 distinct ways to pair them up:
1. $(M_1, W_3)$ vs $(M_2, W_4)$
2. $(M_1, W_4)$ vs $(M_2, W_3)$

This factor of 2 is crucial. It represents the internal arrangements of the match. Without it, we would be undercounting the total number of possible matches.

### Phase 4: The Algebraic Climax

Now, we combine our logic into one elegant equation. The total number of matches is the product of our selections and our arrangements:

$$\binom{n}{2} 	imes \binom{n-2}{2} 	imes 2 = 840$$

Let us expand these combinations. Recall that $\binom{n}{2} = \frac{n(n-1)}{2}$. Our equation becomes:

$$\frac{n(n-1)}{2} 	imes \frac{(n-2)(n-3)}{2} 	imes 2 = 840$$

Watch the magic happen. One of the 2s in the denominator cancels out with the 2 in the numerator. We are left with:

$$\frac{n(n-1)(n-2)(n-3)}{2} = 840$$

Cross-multiplying by 2, we get:

$$n(n-1)(n-2)(n-3) = 1680$$

Here is where many students panic and try to expand this into a quartic equation. Do not do that! We are looking for the product of 4 consecutive integers that equals 1680.

Factoring 1680, we find:

$$1680 = 8 	imes 7 	imes 6 	imes 5$$

Comparing $n(n-1)(n-2)(n-3)$ with $8 	imes 7 	imes 6 	imes 5$, we immediately see that $n=8$.

### The Final Celebration

We have found $n=8$, which is the number of couples. The question asks for the total number of persons.

Since each couple has 2 people, the total number of participants is:

$$2n = 2 	imes 8 = \mathbf{16}$$

We have navigated the constraints, performed the selections, accounted for the arrangements, and solved the algebra with elegance. This is the essence of JEE mathematics—not just calculation, but the art of logical storytelling. Well done!

Animated Solution for Mathematics - Permutations and Combinations: Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ________.

Enter Numerical Value:

Visualized Solution

Defining the Variables

Understanding the Constraint

Selecting the Men

Selecting the Women

Arranging the Match

Forming the Equation

Expanding the Equation

Simplifying the Expression

Solving for $n$

Final Answer

The Sigma Insight: Combinations and Selection

The Dance of the Couples

Phase 1

Phase 2

Phase 3

Phase 4

The Final Celebration

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