A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them

Q: Wait, why can't we just pick 3 ladies from the total pool of 7 ladies and 3 men from the total pool of 7 men?

That is a classic trap! If you just pick 3 ladies from the total pool of 7, you ignore the constraint that X and Y must each invite exactly 3 of their own friends. By picking from the total pool, you might accidentally pick 3 ladies from X's friends and 0 from Y's, which violates the rule that each person must contribute 3 guests. We must respect the individual constraints of X and Y first.

Q: Why are the cases mutually exclusive? Can't we have a mix?

Think of the variable $x_L$ (the number of ladies X invites) as a state of the system. In any single party configuration, X either invites 0, 1, 2, or 3 ladies. He cannot invite both 1 lady and 2 ladies at the same time. Because these scenarios represent distinct, non-overlapping states of the guest list, they are mutually exclusive, and we must add their probabilities (or in this case, the number of ways) together.

Q: Is there a faster way to solve this using generating functions?

Yes, you could use the coefficient of $x^3y^3$ in the expansion of a polynomial, but for a problem of this size, the case-by-case method is actually faster and much less prone to algebraic errors. It keeps you grounded in the physical reality of the party, which is the best way to avoid silly mistakes in JEE Advanced.

Question

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is:

Options:

(A) 485

(B) 468

(C) 469

(D) 484

Sigma Percentile · Accepted Answer

### The Art of Combinatorial Harmony

Welcome, future engineer. Today, we are not just solving a counting problem; we are orchestrating a party. Imagine you are the event planner for Man $X$ and his wife, $Y$.

They have distinct circles of friends, and they want to host a gathering that is perfectly balanced. This is a classic JEE Advanced problem that tests your ability to manage constraints and visualize dependencies. Let us break this down.

### Phase 1: The Constraints

First, let us look at our resources. Man $X$ has 4 ladies and 3 men in his circle. Wife $Y$ has 3 ladies and 4 men.

They have no common friends, which is a blessing—it means our choices are independent. The rules are strict: they must invite 6 guests total, with each spouse contributing exactly 3 friends.

Furthermore, the final guest list must contain exactly 3 ladies and 3 men. This is our **Gender Balance** constraint.

### Phase 2: The Logic of Dependency

Here is the secret to unlocking this problem: dependency. Let $x_L$ be the number of ladies Man $X$ invites.

Since $X$ must invite 3 friends in total, he must invite $(3 - x_L)$ men. Now, look at the total requirement. We need 3 ladies in total.

If $X$ brings $x_L$ ladies, Wife $Y$ must bring $(3 - x_L)$ ladies to reach the total of 3. Consequently, if $Y$ brings $(3 - x_L)$ ladies, she must bring $x_L$ men to complete her quota of 3 friends.

This simple algebraic link, $x_L \in \{0, 1, 2, 3\}$, defines our four distinct cases.

### Phase 3: The Four Acts

Let us walk through each scenario.

**Case 1: $x_L = 3$ (The All-Lady Choice for $X$)**

If $X$ invites 3 ladies, he invites 0 men. The number of ways is:
$$^4C_3 	imes ^3C_0 = 4 	imes 1 = 4$$

To balance, $Y$ must invite 0 ladies and 3 men. The number of ways is:
$$^3C_0 	imes ^4C_3 = 1 	imes 4 = 4$$

Multiplying these, we get $4 	imes 4 = 16$ ways.

**Case 2: $x_L = 2$ (The Balanced Choice)**

If $X$ invites 2 ladies, he must invite 1 man. The ways are:
$$^4C_2 	imes ^3C_1 = 6 	imes 3 = 18$$

$Y$ must invite 1 lady and 2 men. The ways are:
$$^3C_1 	imes ^4C_2 = 3 	imes 6 = 18$$

Multiplying these, we get $18 	imes 18 = 324$ ways. This is our most significant contributor!

**Case 3: $x_L = 1$ (The Inverse Balanced Choice)**

If $X$ invites 1 lady, he must invite 2 men. The ways are:
$$^4C_1 	imes ^3C_2 = 4 	imes 3 = 12$$

$Y$ must invite 2 ladies and 1 man. The ways are:
$$^3C_2 	imes ^4C_1 = 3 	imes 4 = 12$$

Multiplying these, we get $12 	imes 12 = 144$ ways.

**Case 4: $x_L = 0$ (The All-Man Choice for $X$)**

If $X$ invites 0 ladies, he invites 3 men. The ways are:
$$^4C_0 	imes ^3C_3 = 1 	imes 1 = 1$$

$Y$ must invite 3 ladies and 0 men. The ways are:
$$^3C_3 	imes ^4C_0 = 1 	imes 1 = 1$$

Multiplying these, we get $1 	imes 1 = 1$ way.

### Phase 4: The Grand Summation

Now, we bring it all together. Because these cases are mutually exclusive—$X$ cannot invite both 2 ladies and 1 lady simultaneously—we add the results:

$$16 + 324 + 144 + 1 = 485$$

There you have it. **485 ways** to host the perfect party. The beauty of this problem lies in how the constraints force the variables to dance in harmony. Keep this systematic approach in your toolkit, and no combinatorics problem will ever intimidate you again.

Select Answer:

Visualized Solution

Problem Breakdown and Constraints

Party Constraints

Establishing the Selection Logic

Case 1: X invites 3 Ladies

Case 2: X invites 2 Ladies

Case 3: X invites 1 Lady

Case 4: X invites 0 Ladies

Final Summation

The Sigma Insight: Combinations and Selection

The Art of Combinatorial Harmony

Phase 1

Phase 2

Phase 3

Phase 4

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