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 the party, is : 469 (2) 484 (3) 485 (4) 468

Options

A. 484
B. 485
C. 468
D. 469

Explanation

To solve the problem, we need to find the total number of ways in which X and Y can invite 3 ladies and 3 men to their party, ensuring that 3 friends of X and 3 friends of Y are invited.

### Step-by-Step Solution:

1. **Identify the Friends of X and Y:**
- X has 4 ladies and 3 men.
- Y has 3 ladies and 4 men.

2. **Understand the Invitation Requirement:**
- We need to invite a total of 3 ladies and 3 men.
- Each must invite 3 friends, which means we need to consider different combinations of ladies and men from both X and Y.

3. **Case Analysis:**
We will analyze the different cases based on how many ladies are invited from X and Y.

**Case 1:**
- X invites 3 ladies, Y invites 0 ladies.
- X invites 0 men, Y invites 3 men.
- Combinations:
- Ladies: \( \binom{4}{3} \times \binom{3}{0} \)
- Men: \( \binom{3}{0} \times \binom{4}{3} \)

**Case 2:**
- X invites 2 ladies, Y invites 1 lady.
- X invites 1 man, Y invites 2 men.
- Combinations:
- Ladies: \( \binom{4}{2} \times \binom{3}{1} \)
- Men: \( \binom{3}{1} \times \binom{4}{2} \)

**Case 3:**
- X invites 1 lady, Y invites 2 ladies.
- X invites 2 men, Y invites 1 man.
- Combinations:
- Ladies: \( \binom{4}{1} \times \binom{3}{2} \)
- Men: \( \binom{3}{2} \times \binom{4}{1} \)

**Case 4:**
- X invites 0 ladies, Y invites 3 ladies.
- X invites 3 men, Y invites 0 men.
- Combinations:
- Ladies: \( \binom{4}{0} \times \binom{3}{3} \)
- Men: \( \binom{3}{3} \times \binom{4}{0} \)

4. **Calculate Each Case:**
- **Case 1:**
\[ \binom{4}{3} \times \binom{3}{0} \times \binom{3}{0} \times \binom{4}{3} = 4 \times 1 \times 1 \times 4 = 16 \]

- **Case 2:**
\[ \binom{4}{2} \times \binom{3}{1} \times \binom{3}{1} \times \binom{4}{2} = 6 \times 3 \times 3 \times 6 = 324 \]

- **Case 3:**
\[ \binom{4}{1} \times \binom{3}{2} \times \binom{3}{2} \times \binom{4}{1} = 4 \times 3 \times 3 \times 4 = 144 \]

- **Case 4:**
\[ \binom{4}{0} \times \binom{3}{3} \times \binom{3}{3} \times \binom{4}{0} = 1 \times 1 \times 1 \times 1 = 1 \]

5. **Total Combinations:**
Adding all the cases together:
\[ 16 + 324 + 144 + 1 = 485 \]

### Final Answer:
The total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, ensuring that 3 friends of each are in the party, is **485**.