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Grade: 12

                        

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?

3 years ago

Answers : (2)

Shubham Pitroda
41 Points
							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
						
3 years ago
Aryan
11 Points
							
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.
2 years ago
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