circular arrangements probability

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Fri Jan 10, 2014 11:40 pm

Hello,

Can you please assist with the following:

1) In how many different arrangements can five students stand on a circle? The OA is 24

I was thinking that it should be 120.

2) In how many different arrangements can six trees be planted on the circumference of a circular garden if two arrangements are considered different when the positions of the trees are different relative to those of the others?

OA: 120

Can you please help with these?

Thanks,
Sri

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Source: Beat The GMAT — Problem Solving | Fri Jan 10, 2014 11:40 pm

Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sat Jan 11, 2014 12:06 am

gmattesttaker2 wrote:Hello,

Can you please assist with the following:

1) In how many different arrangements can five students stand on a circle? The OA is 24

I was thinking that it should be 120.
Can you please help with these?

Thanks,
Sri

Hi Sri,

Arranging objects in a circle
There are (n-1)! ways to arrange n distinct objects in circle.
so here N=5
There must be 4! ways to arrange them = 24
hence Answer is 24

Regards,
Uva.

Known is a drop Unknown is an Ocean

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Sat Jan 11, 2014 12:08 am

gmattesttaker2 wrote:Hello,

Can you please assist with the following:

2) In how many different arrangements can six trees be planted on the circumference of a circular garden if two arrangements are considered different when the positions of the trees are different relative to those of the others?

OA: 120

Can you please help with these?

Thanks,
Sri

Sri,

As I mentioned above apply the same logic to this one too
n=6
so 5! ways = 120
answer is 120

Regards,
Uva

Known is a drop Unknown is an Ocean

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Jan 11, 2014 3:47 am

To count CIRCULAR arrangements:

1. Place one element in the circle.
2. Count the number of ways to arrange the REMAINING elements.

In how many different arrangements can five students stand on a circle?

Once the first student has been placed, the number of ways to arrange the remaining 4 students relative to the first student = 4! = 24.

In how many different arrangements can six trees be planted on the circumference of a circular garden if two arrangements are considered different when the positions of the trees are different relative to those of the others?

Once the first tree has been placed, the number of ways to arrange the remaining 5 trees relative to the first tree = 5! = 120.

Last edited by GMATGuruNY on Sun Jun 12, 2016 6:29 am, edited 1 time in total.

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Sat Jan 11, 2014 12:02 pm

GMATGuruNY wrote:To count CIRCULAR arrangements:

1. Place one element in the circle.
2. Count the number of ways to arrangement the REMAINING elements RELATIVE to the first element.

In how many different arrangements can five students stand on a circle?
Once the first student has been placed, the number of ways to arrange the remaining 4 students relative to the first student = 4! = 24.

In how many different arrangements can six trees be planted on the circumference of a circular garden if two arrangements are considered different when the positions of the trees are different relative to those of the others?
Once the first tree has been placed, the number of ways to arrange the remaining 5 trees relative to the first tree = 5! = 120.

Hi Mitch,

Thanks for your detailed explanation. I was just wondering if my following understanding is correct here. I have 3 elements A, B and C that I need to arrange in a Circular arrangement. I arrange them as follows. I was not sure if it also matters that the elements have the same order clockwise and anti-clockwise.

Thanks a lot for your help,
Sri

Attachments

: Circular arrangements

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Jan 12, 2014 4:50 am

gmattesttaker2 wrote:
Hi Mitch,

Thanks for your detailed explanation. I was just wondering if my following understanding is correct here. I have 3 elements A, B and C that I need to arrange in a Circular arrangement. I arrange them as follows. I was not sure if it also matters that the elements have the same order clockwise and anti-clockwise.

Thanks a lot for your help,
Sri

i, iv and vi are DUPLICATES.
In each case, the clockwise ordering is the SAME:
A-B-C.

ii, iii and v are also duplicates.
In each case, the clockwise ordering is the same:
A-C-B.

Hence, there are only 2 ways to arrange A, B and C around a circular table:
Once A has been placed, the number of ways to arrange B and C = 2! = 2.

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

by gmattesttaker2 » Tue Jan 14, 2014 7:26 pm

GMATGuruNY wrote:
gmattesttaker2 wrote:
Hi Mitch,

Thanks for your detailed explanation. I was just wondering if my following understanding is correct here. I have 3 elements A, B and C that I need to arrange in a Circular arrangement. I arrange them as follows. I was not sure if it also matters that the elements have the same order clockwise and anti-clockwise.

Thanks a lot for your help,
Sri

i, iv and vi are DUPLICATES.
In each case, the clockwise ordering is the SAME:
A-B-C.

ii, iii and v are also duplicates.
In each case, the clockwise ordering is the same:
A-C-B.

Hence, there are only 2 ways to arrange A, B and C around a circular table:
Once A has been placed, the number of ways to arrange B and C = 2! = 2.

Hello Mitch,

Thank you very much for the explanation.

Best Regards,
Sri

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