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28 Feb 2013, 12:24
Kudos
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B
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:
32% (01:03) correct
68%
(01:07)
wrong
based on 240
sessions
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In how many different ways (relative to each other) can 8 friends sit around a square table with 2 seats on each side of the table?
A. \(2 * 8!\)
B. \(2 * 7!\)
C. \(\frac{8!}{4!}\)
D. \((8-1)!\)
E. \(\frac{8!}{2}\)
A. \(2 * 8!\)
B. \(2 * 7!\)
C. \(\frac{8!}{4!}\)
D. \((8-1)!\)
E. \(\frac{8!}{2}\)
28 Feb 2013, 13:44
Kudos
Bookmarks
Normally the formula for items in a circle is (n-1)! - in this case this is the 7!
It is multiplied by 2 because of the rest of the information. If you have the same arrangement of people you can have everyone shift one position to the right (or left) and they will be sitting next to a different person. With 2 shifts they will be back to sitting next to the original person.
Hope this helps.
It is multiplied by 2 because of the rest of the information. If you have the same arrangement of people you can have everyone shift one position to the right (or left) and they will be sitting next to a different person. With 2 shifts they will be back to sitting next to the original person.
Hope this helps.
24 Jun 2015, 21:13
Kudos
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mun23
I gave the solution to this problem in the post after the post on circular arrangements:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... ts-part-i/
The answer is 2*7!
It's explained in detail on the link given above.
