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Bunuel
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Eighteen guests have to be seated, half on each side a long table. Fou 03 Dec 2021, 08:45
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57% (02:16) correct 43% (02:38) wrong based on 152 sessions

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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) \(11C5 * 9!\)



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B
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Eighteen guests have to be seated, half on each side a long table. Fou 04 Dec 2021, 06:59
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Bunuel
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
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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
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Eighteen guests have to be seated, half on each side a long table. Fou 04 Dec 2021, 07:31
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Bunuel
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) \(11C5 * 9!\)



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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
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Eighteen guests have to be seated, half on each side a long table. Fou 09 Sep 2023, 11:01
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chetan2u
Bunuel
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


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
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chetan2u Bunuel KarishmaB

There were two 11C5 (one for each side), then why you have taken only 1?? Experts Any alternative solution please.
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Eighteen guests have to be seated, half on each side a long table. Fou 10 Sep 2023, 11:13
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chetan2u
Bunuel
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


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

chetan2u Bunuel KarishmaB

There were two 11C5 (one for each side), then why you have taken only 1?? Experts Any alternative solution please.
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When you take 11C5, it gives you two groups one of 5 and other of 6.
So if we calculate 11C5 for A and 11C6 for B, you are taking each combination twice.
Say the numbers are 1 to 11 and 1,2,3,4,5 are placed on table 1. The remaining six, 6,7,8,9,10,11 automatically move to table B.
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Eighteen guests have to be seated, half on each side a long table. Fou 26 Nov 2025, 10:06
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