1. Interpret the condition carefully

We need a party of 6 people consisting of:

• $3$ ladies
• $3$ men

Also, among these $6$ invitees:

• exactly $3$ must be friends of $X$
• exactly $3$ must be friends of $Y$

Since $X$ and $Y$ have no common friends, this means we choose:

• $3$ people from $X$'s friends
• $3$ people from $Y$'s friends

such that total ladies = 3 and total men = 3.

2. Friend distribution

• $X$ has $4$ lady friends and $3$ men friends.
• $Y$ has $3$ lady friends and $4$ men friends.

Suppose from $X$'s side we choose:

• $k$ ladies, so men chosen from $X$ = $3-k$

Then from $Y$'s side, since total must be $3$ ladies and $3$ men:

• ladies from $Y$ = $3-k$
• men from $Y$ = $k$

So $k$ can take values such that all selections are possible.

3. Count for each possible $k$

Case 1: $k=0$

From $X$: $0$ ladies, $3$ men

Number of ways: $\binom{4}{0}\binom{3}{3}=1$

From $Y$: $3$ ladies, $0$ men

Number of ways: $\binom{3}{3}\binom{4}{0}=1$

Total ways in this case: $1\cdot 1=1$

Case 2: $k=1$

From $X$: $1$ lady, $2$ men

Ways: $\binom{4}{1}\binom{3}{2}=4\cdot 3=12$

From $Y$: $2$ ladies, $1$ man

Ways: $\binom{3}{2}\binom{4}{1}=3\cdot 4=12$

Total: $12\cdot 12=144$

Case 3: $k=2$

From $X$: $2$ ladies, $1$ man

Ways: $\binom{4}{2}\binom{3}{1}=6\cdot 3=18$

From $Y$: $1$ lady, $2$ men

Ways: $\binom{3}{1}\binom{4}{2}=3\cdot 6=18$

Total: $18\cdot 18=324$

Case 4: $k=3$

From $X$: $3$ ladies, $0$ men

Ways: $\binom{4}{3}\binom{3}{0}=4\cdot 1=4$

From $Y$: $0$ ladies, $3$ men

Ways: $\binom{3}{0}\binom{4}{3}=1\cdot 4=4$

Total: $4\cdot 4=16$

4. Add all cases

$1+144+324+16=485$

5. Compare with options

The total number of ways is: $\boxed{485}$

So the correct option is: D: 485

6. Comparison with stored answer

Stored correct answer = D

Our derived answer = D

So they agree.