Question : 18 guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is 11! ( 9!) ( 9! ) / 5! 6!.
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rest guests = 18 -7 = 11
five seats left in one side and 6 other side
as 9 on one side is there which is occupied by 4 and 3 guests respectively
So select five from 11
11C5
and now 6 left and we have to select 6
So 6C6
Arrange both side guests
9! × 9!
so no of ways = 11C5 × 6C6 × 9!× 9!
= 11!× 9!× 9!/5! 6!
Yeah it's correct
five seats left in one side and 6 other side
as 9 on one side is there which is occupied by 4 and 3 guests respectively
So select five from 11
11C5
and now 6 left and we have to select 6
So 6C6
Arrange both side guests
9! × 9!
so no of ways = 11C5 × 6C6 × 9!× 9!
= 11!× 9!× 9!/5! 6!
Yeah it's correct
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