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03 Dec 2021, 08:45
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:16) correct
43%
(02:38)
wrong
based on 152
sessions
History
Date
Time
Result
Not Attempted Yet
Eighteen guests have to be seated, half on each side a long table. Four particular guest desires to sit on one particular side and three others on the other side. The number of ways in which the seating arrangement can be made, is
(A) \(9! *9!\)
(B) \(11C5 * 9! * 9!\)
(C) \((\frac{11!}{5!}) * 9! * 9!\)
(D) \(11C5\)
(E) \(11C5 * 9!\)
Are You Up For the Challenge: 700 Level Questions
(A) \(9! *9!\)
(B) \(11C5 * 9! * 9!\)
(C) \((\frac{11!}{5!}) * 9! * 9!\)
(D) \(11C5\)
(E) \(11C5 * 9!\)
Are You Up For the Challenge: 700 Level Questions
04 Dec 2021, 06:59
Kudos
Bookmarks
Bunuel
Let the two sides of table be A, where 4 are to seated, and B, where 3 are to be seated.
A: 4 are already seated
The remaining 5 can be chosen from 11 in 11C5.
B: 3 are already seated
The remaining 6 are already segregated above in 11C5.
Finally, 9 on each side can be arranged in 9! ways.
Total = 11C5*9!*9!
B
04 Dec 2021, 07:31
Kudos
Bookmarks
Bunuel
let us first place the 4 and 3 people in their prefereed side
remaning people to be chosen is 11C5
Then arranging each side in 9!*9! ways
Therefore total possibilities =11C5*9!*9!
Therefore IMO B
