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GMAT Prep Exam 1- Arranging Group

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GMAT Prep Exam 1- Arranging Group

by nsuen » Sun Jun 12, 2016 5:37 am
Hi-

Q: At a dinner party, 5 people are to be seated around a circular table, two seating arrangement are considered different only when the the position of the people are different relative to each other. What is the total number of different possible seating arrangement for the group?

My thinking process:

5! /( 3! 2!) because total are 5! ( so put in numerator, then because the position of the people need to be different to count , so I drew up a table and see that after A seats on seat 1, it has to go to seat 3, otherwise, it wont count as different way ( I dont know does it make sense? )
My answer was 10, but the correct answer is 24. Could someone explain to me why or is there a better way to do this?

Thanks!
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by GMATGuruNY » Sun Jun 12, 2016 6:30 am
To count CIRCULAR arrangements:

1. Place one element in the circle.
2. Count the number of ways to arrange the REMAINING elements.
nsuen wrote:At a dinner party, 5 people are to be seated around a circular table, two seating arrangement are considered different only when the the position of the people are different relative to each other. What is the total number of different possible seating arrangement for the group?
After the first person has been seated, the number of ways to arrange the remaining 4 people = 4! = 24.
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by [email protected] » Sun Jun 12, 2016 10:27 am
Hi nsuen,

The "math" behind Permutation questions will be influenced by the restrictions that the prompt gives you to work with, so you have to pay careful attention to how these questions are worded.

Here, we know that there are 5 people and that they're sitting around a circular table. This is different from 5 chairs that are in a row (which would be a much easier calculation).

With 5 seats in a row, we'd have 5x4x3x2x1 = 120 arrangements....BUT since we have a circular table, some of these arrangements are duplicates. With 5 chairs, there are 5 "starting spots" that would lead to duplicates. For example...

ABCDE
BCDEA
CDEAB
DEABC
EABCD

These 5 options are actually the SAME option, since the letters are sill in order "going around" the table. We can't count this as 5 options; it's just 1 option.

So the math becomes: 5!/5 = 5x4x3x2x1/5 = 24

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by nsuen » Sun Jun 12, 2016 8:07 pm
Hi Rich:

I am wondering why do we divide by 5? Thanks!
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by [email protected] » Mon Jun 13, 2016 8:39 am
Hi nsuen,

Since you have 5 people sitting at a circular table, you divide by 5 because rotating the people in either direction does NOT create a new arrangement.

Try drawing a circle, then write A, B, C, D and E on it (representing the 5 people). If you rotate the drawing, it might look like the people are in different spots, but the arrangement does NOT change. The 5 people sit in the same arrangement (even if it's oriented differently), but you're not allowed to count that arrangement 5 times - you only get to count it ONCE. Thus, you have to divide the total by 5.

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