Question

7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite in a dinner party, 3 ladies and 3 gentlemen so that there are 3 of man’s relatives and 3 of wife’s relatives? 

(a) 265 

(b) 375 

(c) 395 

(d) 485

Answer 1

(d) 485

The 6 people (3 ladies and 3 gentlemen) to be invited can be selected in the given ways: 

(i) 3 ladies from man’s relatives and 3 gentlemen from wife’s relatives, i.e., 

No. of ways = ⁴C₃ × ³C₀ × ³C₀ × ⁴C₃ = 16 

(ii) 2 ladies and 1 gentleman from man’s relatives and 1 lady and 2 gentlemen from wife’s relatives, i.e., 

No. of ways = ⁴C₂ × ³C₁ × ³C₁ × ⁴C₂ 

= \(\frac{4\times3}{2}\times3\times3\times\frac{4\times3}{2}\) = 324

(iii) 1 lady and 2 gentlemen from man’s relatives and 2 ladies and 1 gentleman from wife’s relatives, i.e., 

No. of ways = ⁴C₁ × ³C₂ × ³C₂ × ⁴C₁ = 4 × 3 × 4 × 3 = 144 

(iv) 3 gentlemen from man’s relatives and 3 ladies from wife’s relatives, i.e., 

No. of ways = ⁴C₀ × ³C₃ × ³C₃ × ⁴C₀ = 1 × 1 = 1 

∴ Total number of ways of selecting the relatives for dinner party = 16 + 324 + 144 + 1 = 485.