Key Concepts and Formulas

• Combinations ($\binom{n}{k}$): The number of ways to choose $k$ items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

• Product Rule (Fundamental Principle of Multiplication): If an event can occur in $m$ ways and an independent second event can occur in $n$ ways, then the two events can occur in $m \times n$ ways.

• Sum Rule (Fundamental Principle of Addition): If an event can occur in $m$ ways and a separate, mutually exclusive event can occur in $n$ ways, then one of these events can occur in $m + n$ ways.

Step-by-Step Solution

Step 1: Define Variables and Constraints

• $L_X$: Number of ladies among X's friends (4)
• $M_X$: Number of men among X's friends (3)
• $L_Y$: Number of ladies among Y's friends (3)
• $M_Y$: Number of men among Y's friends (4)
• $l_X$: Number of ladies invited from X's friends
• $m_X$: Number of men invited from X's friends
• $l_Y$: Number of ladies invited from Y's friends
• $m_Y$: Number of men invited from Y's friends

Constraints:

• $l_X + l_Y = 3$ (Total ladies invited)
• $m_X + m_Y = 3$ (Total men invited)
• $l_X + m_X = 3$ (X invites 3 friends)
• $l_Y + m_Y = 3$ (Y invites 3 friends)
• $0 \le l_X \le 4$, $0 \le m_X \le 3$, $0 \le l_Y \le 3$, $0 \le m_Y \le 4$

Step 2: Apply Implicit Constraint

Assume X and Y each invite at least one lady and one man. This means:

• $l_X \ge 1$ and $m_X \ge 1$
• $l_Y \ge 1$ and $m_Y \ge 1$

Step 3: Enumerate Possible Cases based on X's invitees

Since $l_X + m_X = 3$, and $l_X \ge 1$ and $m_X \ge 1$, the possible values for $l_X$ are 1 and 2.

Case 1: $l_X = 1$ and $m_X = 2$

• X invites 1 lady and 2 men.
• $l_Y = 3 - l_X = 3 - 1 = 2$
• $m_Y = 3 - m_X = 3 - 2 = 1$
• Y invites 2 ladies and 1 man.

Step 4: Calculate the number of ways for Case 1

• Ways for X to invite 1 lady and 2 men: $\binom{4}{1} \times \binom{3}{2} = 4 \times 3 = 12$
• Ways for Y to invite 2 ladies and 1 man: $\binom{3}{2} \times \binom{4}{1} = 3 \times 4 = 12$
• Total ways for Case 1: $12 \times 12 = 144$

Step 5: Case 2: $l_X = 2$ and $m_X = 1$

• X invites 2 ladies and 1 man.
• $l_Y = 3 - l_X = 3 - 2 = 1$
• $m_Y = 3 - m_X = 3 - 1 = 2$
• Y invites 1 lady and 2 men.

Step 6: Calculate the number of ways for Case 2

• Ways for X to invite 2 ladies and 1 man: $\binom{4}{2} \times \binom{3}{1} = 6 \times 3 = 18$
• Ways for Y to invite 1 lady and 2 men: $\binom{3}{1} \times \binom{4}{2} = 3 \times 6 = 18$
• Total ways for Case 2: $18 \times 18 = 324$

Step 7: Calculate the Total Number of Ways

• Total ways = Ways for Case 1 + Ways for Case 2
• Total ways = $144 + 324 = 468$

Common Mistakes & Tips

• Forgetting the implicit constraint: Problems involving parties or group formations might implicitly require a mix of genders if possible. Always check if adding such a constraint aligns with the answer choices.
• Incorrectly calculating combinations: Double-check your combination calculations to avoid arithmetic errors.
• Not considering all possible cases: Ensure you've exhausted all valid cases based on the given and implicit constraints.

Summary

By considering the number of ladies and men invited by X and Y, along with the implicit constraint that each invites at least one of each gender, we systematically analyzed the possible cases and calculated the total number of ways they can throw the party. The total number of ways is 468.

Final Answer

The final answer is $\boxed{468}$, which corresponds to option (A).