Last visit was: 19 Nov 2025, 18:11 It is currently 19 Nov 2025, 18:11
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,372
 [7]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,372
 [7]
Eighteen guests have to be seated half on each side of a long table. 23 Feb 2021, 06:30
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

70% (01:52) correct 30% (02:14) wrong based on 176 sessions

History

Date
Time
Result
Not Attempted Yet
Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit one particular side and three other on the other side. Determine the number of ways in which the sitting arrangements can be made.

A. 9P4 × 9P3 × (9)!
B. 9P4 × 9P3 × (10)!
C. 9P4 × 9P3 × (11)!
D. 9P4 × 9P3 × (12)!
E. 9P4 × 9P3 × (13)!

Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
ShowHide Answer
Official Answer
C
User avatar
Lipun
User avatar
Joined: 05 Jan 2020
Last visit: 08 Jan 2025
Posts: 144
Own Kudos:
157
 [1]
Given Kudos: 291
Posts: 144
Kudos: 157
 [1]
Eighteen guests have to be seated half on each side of a long table. 23 Feb 2021, 10:07
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Number of ways to choose 5 people out of 11 = 11C5
Number of arrangements on each side of the table = 9!

Total possible arrangements = 11C5 * 9! * 9! = (11!/5!*6!)*9!*9! = 11! * (9!/5!) * (9!/6!) = 11! * 9P4 * 9P3
User avatar
Mona2019
User avatar
Joined: 13 Mar 2018
Last visit: 27 Sep 2022
Posts: 190
Own Kudos:
218
 [1]
Given Kudos: 1,608
Posts: 190
Kudos: 218
 [1]
Eighteen guests have to be seated half on each side of a long table. 02 Mar 2021, 10:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel , chetan2u
Could you please help me with this question.
How it is 12! , I am not able to get it.

Thanks in advance for your help.

Posted from my mobile device
User avatar
TestPrepUnlimited
User avatar
Joined: 17 Sep 2014
Last visit: 30 Jun 2022
Posts: 1,224
Own Kudos:
1,111
 [1]
Given Kudos: 6
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Expert
Expert reply
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Posts: 1,224
Kudos: 1,111
 [1]
Eighteen guests have to be seated half on each side of a long table. 02 Mar 2021, 13:37
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Mona2019 it looks like the answer choices have a typo, it should be 11! for C and 12! for D etc.

My solution:

Slot in the four guests first, they have 9 seats to choose from so that is 9P4 options (9*8*7*6).
The next three guests on the other side of the table have 9 seats to choose from as well, that is 9P3 options for them.

Finally there are 11 seats to fill in with 11 guests, so 11P11 or 11! for those 11 people.

Since these three groups of guests are independent, we multiply all of the above to get 9P4 * 9P3 * 11!.

Ans: C
User avatar
Mona2019
User avatar
Joined: 13 Mar 2018
Last visit: 27 Sep 2022
Posts: 190
Own Kudos:
218
Given Kudos: 1,608
Posts: 190
Kudos: 218
Eighteen guests have to be seated half on each side of a long table. 02 Mar 2021, 22:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I thought I must be missing something .

Thank you for replying and clearing my doubt . TestPrepUnlimited

Posted from my mobile device
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 19 Nov 2025
Posts: 21,716
Own Kudos:
26,997
Given Kudos: 300
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 21,716
Kudos: 26,997
Eighteen guests have to be seated half on each side of a long table. 13 Mar 2021, 15:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit one particular side and three other on the other side. Determine the number of ways in which the sitting arrangements can be made.

A. 9P4 × 9P3 × (9)!
B. 9P4 × 9P3 × (10)!
C. 9P4 × 9P3 × (11)!
D. 9P4 × 9P3 × (12)!
E. 9P4 × 9P3 × (13)!

Show more
Solution:

There are 9P4 ways for the 4 guests who want to sit on one side of the table and 9P3 ways for the 3 other guests who want to sit on the other side of the table. Since there are no restrictions for the remaining 11 guests, there are 11P11 = 11! ways for them to sit at the remaining seats. Therefore, the total number of seating arrangements is 9P4 x 9P3 x 11!.

Answer: C
User avatar
PSKhore
User avatar
Joined: 28 Apr 2025
Last visit: 19 Nov 2025
Posts: 175
Own Kudos:
27
Given Kudos: 105
Products:
Forum Quiz
Posts: 175
Kudos: 27
Eighteen guests have to be seated half on each side of a long table. 11 Aug 2025, 00:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit one particular side and three other on the other side. Determine the number of ways in which the sitting arrangements can be made.

A. 9P4 × 9P3 × (9)!
B. 9P4 × 9P3 × (10)!
C. 9P4 × 9P3 × (11)!
D. 9P4 × 9P3 × (12)!
E. 9P4 × 9P3 × (13)!

Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Show more
Number of ways of choosing and seating 4 guests out of half of 18 i.e 9 = P(9,4)
Number of ways of choosing and seating 3 guests out of half of 18 i.e 9 = P(9,3)

Remaining guests = (18-7) = 11
These can be arranged in 11! ways

Hence, C
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts