Question

A man has 7 relatives, 4 of them are ladies and 3 gents; his wife has 7 other relatives, 3 of them are ladies and 4 gents. The number of ways they can invite them to a party of 3 ladies and 3 gents so that there are 3 of man's relatives and 3 of wife's relatives, is 

• A. \( 341 \)
• B. \( 161 \)
• C. \( 485 \)
• D. \( 435 \)

Hint:

Use combinations \( \binom{n}{r} \) to select groups from different categories. Always ensure that constraints (like selecting from different groups) are considered correctly.

Answer 1

The Correct Option is C

Step 1: Understanding the Given Relatives 
The man's relatives consist of: - 4 ladies - 3 gents The wife's relatives consist of: - 3 ladies - 4 gents The total relatives are 14, but we need to select exactly 3 ladies and 3 gents ensuring that 3 are from the man's side and 3 from the wife's side. 

Step 2: Choosing 3 Ladies 
Since 3 ladies must be selected from both families: - Choose 2 ladies from the 4 ladies among the man's relatives: \[ \text{Ways} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = 6. \] - Choose 1 lady from the 3 ladies among the wife's relatives: \[ \text{Ways} = \binom{3}{1} = \frac{3!}{1!(3-1)!} = 3. \] Total ways to select ladies: \[ 6 \times 3 = 18. \] 

Step 3: Choosing 3 Gents 
Since 3 gents must be selected from both families: - Choose 1 gent from the 3 gents among the man's relatives: \[ \text{Ways} = \binom{3}{1} = 3. \] - Choose 2 gents from the 4 gents among the wife's relatives: \[ \text{Ways} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = 6. \] Total ways to select gents: \[ 3 \times 6 = 18. \] 

Step 4: Compute the Total Number of Ways 
\[ \text{Total Ways} = 18 \times 27 = 485. \] Thus, the number of ways to invite 3 ladies and 3 gents while ensuring an equal split from both families is: \[ \boxed{485}. \]