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

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Manager
Joined: 09 Feb 2013
Posts: 120
At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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11 Jun 2013, 04:02
3
16
00:00

Difficulty:

25% (medium)

Question Stats:

72% (01:03) correct 28% (01:07) wrong based on 437 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 47101
Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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11 Jun 2013, 05:32
10
15
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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Intern
Joined: 26 Oct 2013
Posts: 23
Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 09:40
1
Hi Bunuel, I find the same problem but with a rectangular table. How should be the reasoning there?

Thks
Intern
Joined: 26 Oct 2013
Posts: 23
Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 09:56
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?
Math Expert
Joined: 02 Sep 2009
Posts: 47101
Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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30 Mar 2014, 11:29
2
5
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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21 May 2018, 07:12
Another way of finding the # of cases where the restriction is violated, i.e. when the two chief of staffs are together. # of cases when the two generals are together.

there are 7 spots. The first general who wants to be next to the other one. He can choose whatever spot (7). The second general that wants to be next to the first has 2 choices (2). The rest of them have their regular amount of choice, given that two spots are now occupied (5!).

Answer = 7*2*5!. If we care about relative positioning rather than absolute positioning, we divide that by 7 and found aforementionned answer.

For n slots, answer = n*2*(n-2)!
Senior Manager
Joined: 31 May 2017
Posts: 285
Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of [#permalink]

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21 May 2018, 22:39
1
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?

Number of ways 7 person can be arranged in a circular table = (N-1)! = (7-1)! = 6! = 720

Lets consider 2 people not sitting together as one entity = so we have 6 people and number of arrangements is 5!. The 2 people can be arranged in 2 ways so probability is 2*5! = 240

Total probability is = 720-240 = 480

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

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24 May 2018, 10:52
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

Recall that the circular permutations formula for n items is (n - 1)!

The number of ways for the Chief of Naval Operations (N) to sit next to Chief of the National Guard Bureau (G) is to treat the pair [N - G] as a single entity. The seating arrangement can be shown as:

[N - G] - A - B - C - D - E

Since we have a circle, that above arrangement can be arranged in (6 -1)! = 5! = 120 ways. Additionally, we could seat N and G as [G - N], giving us an additional (6 -1)! = 120 ways, for a total of 120 + 120 = 240 ways in which the two chiefs would sit next to each other.
The total number of ways for all seating arrangements is (7 - 1)! = 6! = 720.

Thus, the number ways the Chief of Naval Operations does not sit next to the Chief of the National Guard Bureau is 720 - 240 = 480.

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Re: At a meeting of the 7 Joint Chiefs of Staff, the Chief of   [#permalink] 24 May 2018, 10:52
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# At a meeting of the 7 Joint Chiefs of Staff, the Chief of

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