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19 Nov 2020, 09:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:38) correct
47%
(02:31)
wrong
based on 177
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At a dinner party, 8 students (Hermione, Ron, Fred, George, Neville, and 3 Beauxbatons students) are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group if the 3 Beauxbatons students insist on sitting next to each other and Hermione and Ron insist on sitting apart?
A. 36
B. 72
C. 108
D. 432
E. 1,728
A. 36
B. 72
C. 108
D. 432
E. 1,728
ananya3
Current Student
Joined: 05 Oct 2018
Last visit: 17 Dec 2022
Posts: 106
Given Kudos: 152
Location: India
Schools: Fuqua '24 Haas '24 ISB '23 Johnson '24 Kelley '24 Kenan-Flagler '24 McDonough '24 NUS Ross '24 Tepper '24 Tuck '24 Rotman '24 (A)
GMAT 1: 700 Q49 V36
GPA: 3.87
Schools: Fuqua '24 Haas '24 ISB '23 Johnson '24 Kelley '24 Kenan-Flagler '24 McDonough '24 NUS Ross '24 Tepper '24 Tuck '24 Rotman '24 (A)
GMAT 1: 700 Q49 V36
Posts: 106
19 Nov 2020, 20:00
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Bookmarks
IMO D
If we sit 8 different students around a circular table
number of possible arrangements = (8-1)! = 7! (around a circular table fix any 1 position for a person - 1 way and then arrange the remaining 7 people in 7! ways )
but the 3 Beauxbatons students want to sit together
=> we need to arrange {Hermione, Ron, Fred, George, Neville, {B1,B2,B3}} (6 entities) around the table
=> arrangement for 6 entities around the circular table = (6-1)! = 5!
Also, the 3 Beauxbatons can be arranged in 3! ways among themselves
=> total number of arrangements = 5! * 3! = 720
Hermione and Ron cannot be sitting together, so subtract the combination when they are sitting next to each other from the total
So we have ({{Hermione, Ron}, Fred, George, Neville, {B1,B2,B3}}) 5 possible entities when Ron and Hermione are sitting together
=> Number possible of arrangements = 4! * 2! * 3! (4! arrangements for 5 entities * 2! possible ways to arrange Ron and Hermione when they are sitting next to each other * 3! ways to arrange the Beauxbatons)
= 288
Therefore, total possible arragements with all the restrictions
= 720 - 288
= 432
If we sit 8 different students around a circular table
number of possible arrangements = (8-1)! = 7! (around a circular table fix any 1 position for a person - 1 way and then arrange the remaining 7 people in 7! ways )
but the 3 Beauxbatons students want to sit together
=> we need to arrange {Hermione, Ron, Fred, George, Neville, {B1,B2,B3}} (6 entities) around the table
=> arrangement for 6 entities around the circular table = (6-1)! = 5!
Also, the 3 Beauxbatons can be arranged in 3! ways among themselves
=> total number of arrangements = 5! * 3! = 720
Hermione and Ron cannot be sitting together, so subtract the combination when they are sitting next to each other from the total
So we have ({{Hermione, Ron}, Fred, George, Neville, {B1,B2,B3}}) 5 possible entities when Ron and Hermione are sitting together
=> Number possible of arrangements = 4! * 2! * 3! (4! arrangements for 5 entities * 2! possible ways to arrange Ron and Hermione when they are sitting next to each other * 3! ways to arrange the Beauxbatons)
= 288
Therefore, total possible arragements with all the restrictions
= 720 - 288
= 432
19 Nov 2020, 11:49
Kudos
Bookmarks
At a dinner party, 8 students (Hermione, Ron, Fred, George, Neville, and 3 Beauxbatons students) are to be seated around a circular table. 2 seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group if the 3 Beauxbatons students insist on sitting next to each other and Hermione and Ron insist on sitting apart?
Arrangement around a circular table = (N-1)!
Total arrangement:
{George}, {Fred}, {Neville}, {Hermione , Ron} , {B1} , {B2} , {B3}
Now we have 7 units and can be arranged in a circular table in (7-1)! = 6! # of ways = 720 ways
Restriction:
1.Hermione and Ron insist on sitting apart.
2.Three Beauxbatons students insist on sitting next to each other.
Arrangement with restriction looks like
{George}, {Fred}, {Neville}, {Hermione , Ron} , {B1, B2, B3}
We have 5 units, so total arrangements = (N-1)! = 4! = 24 ways
Hermione and Ron can be arranged in 2 ways
Beauxbatons can be arranged in 6 ways
Number of arrangements with restriction = 24*2*6 = 288
Total number of arrangements:
The total number of arrangements = {Total} - {Restriction} = 720 - 288 = 432
Ans: D
Arrangement around a circular table = (N-1)!
Total arrangement:
{George}, {Fred}, {Neville}, {Hermione , Ron} , {B1} , {B2} , {B3}
Now we have 7 units and can be arranged in a circular table in (7-1)! = 6! # of ways = 720 ways
Restriction:
1.Hermione and Ron insist on sitting apart.
2.Three Beauxbatons students insist on sitting next to each other.
Arrangement with restriction looks like
{George}, {Fred}, {Neville}, {Hermione , Ron} , {B1, B2, B3}
We have 5 units, so total arrangements = (N-1)! = 4! = 24 ways
Hermione and Ron can be arranged in 2 ways
Beauxbatons can be arranged in 6 ways
Number of arrangements with restriction = 24*2*6 = 288
Total number of arrangements:
The total number of arrangements = {Total} - {Restriction} = 720 - 288 = 432
Ans: D
