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# At a meeting of the 7 Joint Chiefs of Staff, the Chief of

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At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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11 Jun 2013, 03:02
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At a meeting of the 7 Joint Chiefs of Staff, the Chief of Naval Operations does not want to sit next to the Chief of the National Guard Bureau. How many ways can the 7 Chiefs of Staff be seated around a circular table?

A. 120
B. 480
C. 960
D. 2520
E. 5040
[Reveal] Spoiler: OA

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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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11 Jun 2013, 04:32
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emmak wrote:
At a meeting of the 7 Joint Chiefs of Staff, the Chief of Naval Operations does not want to sit next to the Chief of the National Guard Bureau. How many ways can the 7 Chiefs of Staff be seated around a circular table?

A. 120
B. 480
C. 960
D. 2520
E. 5040

7 people can be arranged around circular table in (7-1)!=6! # of ways.

Consider the two people (A and B) who do not want to sit together as one unit: {AB}. Now, 6 units {AB}, {C}, {D}, {E}, {F} and {G} can can be arranged around circular table in (6-1)!=5! # of ways. A and B can be arranged within their unit in two ways {AB} and {BA}. Thus the # of ways those two people sit together is 2*5!.

The number of ways they do not sit together = {Total} - {Restriction} = 6! - 2*5! = 5!(6 - 2) = 480.

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Hope it helps.
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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 08:40
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Hi Bunuel, I find the same problem but with a rectangular table. How should be the reasoning there?

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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 08:56
Bunuel, I'm also a little confused with the number of arrangements of n distinct objects in a circle. Why is it given by (n-1)!. In the veritas answer they say: "answer E (5040), should be the number of ways to arrange all 7 without the seating restriction given". Is this incorrect?
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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 10:29
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GDR29 wrote:
Bunuel, I'm also a little confused with the number of arrangements of n distinct objects in a circle. Why is it given by (n-1)!. In the veritas answer they say: "answer E (5040), should be the number of ways to arrange all 7 without the seating restriction given". Is this incorrect?

I guess you did not follow any of the links given above...

The number of arrangements of n distinct objects in a row is given by $$n!$$.
The number of arrangements of n distinct objects in a circle is given by $$(n-1)!$$.

The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:

$$\frac{n!}{n} = (n-1)!$$.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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26 Apr 2015, 03:42
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03 Jul 2016, 01:30
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