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M30-02

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M30-02  [#permalink]

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New post 16 Sep 2014, 01:45
2
9
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (01:36) correct 45% (01:51) wrong based on 143 sessions

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Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640

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Re M30-02  [#permalink]

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New post 16 Sep 2014, 01:45
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5
Official Solution:

Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640


Consider two brothers and two sisters between them as one unit: {BSSB}.

So, now we have 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}.

These 6 units can be arranged around a circular table in \((6-1)!=5!\) ways.

Next, analyze {BSSB} unit:

We can choose 2 brothers out of 3 for the unit in \(C^2_3=3\) ways;

These brothers, within the unit, can be arranged in 2! ways: \(\{B_1, S, S, B_2\}\) or \(\{B_2, S, S, B_1\}\).

The sisters, within the unit, also can be arranged in 2! ways: \(\{B, S_1, S_2, B\}\) or \(\{B, S_2, S_1, B\}\).

Therefore, the final answer is 5!*3*2*2=1440.


Answer: C
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Re M30-02  [#permalink]

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New post 07 Jul 2016, 10:26
I think this is a high-quality question and I agree with explanation. Hi Bunuel, can you please post some detailed theory on such circular permutations? Thanks
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Re: M30-02  [#permalink]

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New post 07 Jul 2016, 10:35
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Senthil7 wrote:
I think this is a high-quality question and I agree with explanation. Hi Bunuel, can you please post some detailed theory on such circular permutations? Thanks


Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Hope it helps.
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Re: M30-02  [#permalink]

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New post 03 Aug 2016, 13:17
I think this is also possible {b,g,s,g,s,g,g,b},b
"2 sisters are seated between any two of the three brothers"

am I missing anything?
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Re: M30-02  [#permalink]

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New post 27 Nov 2016, 01:33
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No word in the question specifies whether two sisters can be seated jointly or not. The question needs to be rephrased so that it would be clearer.
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Re M30-02  [#permalink]

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New post 19 Oct 2017, 05:01
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I think this is a poor-quality question and I agree with explanation. The question does not specify whether there can be anyone between the two sisters.
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M30-02  [#permalink]

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New post 30 Oct 2018, 11:05
Can someone explain what is wrong with this approach -

Arranging 3 boys and 4 grandparents in circle = (7-1)! = 6! = 720

Selecting one out of 3 slots between 3 boys for 2 sisters = 3C1 = 3

Arranging sisters in that slot = 2! = 2

Total = 720*3*2 = 4320
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Re: M30-02  [#permalink]

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New post 29 Dec 2018, 06:04
These 6 units can be arranged around a circular table in \((6-1)!=5!\) ways.

Can someone explain why its (6-1)! instead of just 6! ?

Thanks!
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Re: M30-02  [#permalink]

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New post 31 Dec 2018, 01:33
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Gruzin wrote:
These 6 units can be arranged around a circular table in \((6-1)!=5!\) ways.

Can someone explain why its (6-1)! instead of just 6! ?

Thanks!


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:

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

Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.
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Re: M30-02   [#permalink] 31 Dec 2018, 01:33
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