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Grade 12Algebra

A MAN HAS 7 RELATIVES,4 OF THEM ARE LADIES AND 3 GENTLEMEN.HIS WIFE HAS ALSO 7 RELATIVES AND 4 GENTLEMEN.IN HOW MANY WAYS CAN THEY INVITE A DINNER PARTY OF 3 LADIES AND 3 GENTLEMEN SO THAT THERE ARE 3 OF MAN’S RELATIVES AND 3 OF THE WIFE’S RELATIVES?

BHUKYA GOPI CHAND

9 Years agoGrade 12

2 Answers

Shubham Pitroda

9 Years ago

There are total 7 relatives in both of mens as well as wives....according to questions we have to select 3 relatives from both of them so 7c3 +7c3 so you will get answer 70....-Shubham Pitroda

Aryan

8 Years ago

There four possibilities:

(i) 3 ladies from husband’s side and 3 gentlemen from wife’s side.

No. of ways in this case

= ⁴ C₃ x ⁴ C₃ = 4 x 4 = 16

(ii) 3 gentlemen from husband’s side and 3 ladies from wife’s side.

No. of ways in this case = ³ C₃ x ³ C ₃ = 1 x 1 = 1

(iii) 2 ladies and one gentleman from husband’s side and lady and 2 gentlemen from wife’s side.

No. of ways in this case

= (⁴ C₄ x ³ C₁) x (³ C₁ x ⁴ C₂) = 6 x 3 x 3 x 6 = 324

(iv) One lady and 2 gentlemen from husband’s side and 2 ladies and one gentlemen from wife’s side.

No. of ways in this case

= (⁴ C₁ x ³ C₂) x (³ C₂ x ⁴ C₁) = 4 x 3 x 3 x 4 = 144

Hence the total no. of ways are

= 16 + 1 + 324 + 144 = 485.

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