JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 19)
Two families with three members each and one family with four members are to be seated in a row.
In how many ways can they be seated so that the same family members are not separated?
2! 3! 4!
(3!)3.(4!)
3! (4!)3
(3!)2.(4!)
Explanation
F1 $$ \to $$ 3 members
F2 $$ \to $$ 3 members
F3 $$ \to $$ 4 members
Total arrangements of three families = 3!
Arrangement between members of F1 family = 3!
Arrangement between members of F2 family = 3!
Arrangement between members of F3 family = 4!
$$ \therefore $$ Total numbers of ways can they be seated so that the same family members are not separated
= 3! $$ \times $$ 3! $$ \times $$ 3! $$ \times $$ 4!
= (3!)3.(4!)
F2 $$ \to $$ 3 members
F3 $$ \to $$ 4 members
Total arrangements of three families = 3!
Arrangement between members of F1 family = 3!
Arrangement between members of F2 family = 3!
Arrangement between members of F3 family = 4!
$$ \therefore $$ Total numbers of ways can they be seated so that the same family members are not separated
= 3! $$ \times $$ 3! $$ \times $$ 3! $$ \times $$ 4!
= (3!)3.(4!)
Comments (0)
