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Nine family members: 5 grandchildren (3 brothers and 2 siste

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Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 25 Jul 2014, 09:09
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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

Kudos for a correct solution.


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Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 25 Jul 2014, 09:10
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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: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 25 Jul 2014, 18:17
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Bunuel wrote:


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

Kudos for a correct solution.



Number of ways to arrange the 2 sisters = \(2\)
Number of ways to choose the 2 brothers out of 3 = \(3C2 = 3\)
Number of ways to arrange the 2 sisters between the 2 brothers (only the brothers switch places since the sisters have to be in the middle) = \(2\)
Number of ways to arrange the 2 sisters between the 2 brothers with the leftover brother and 4 grand parents around a table = \((6-1)! = 5!\)

\(2x3x2x5! = 12*120=1440\)
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 26 Jul 2014, 05:56
What am I doing wrong? I got 480...

So, around the table, we have: {brother, sister, sister, brother}, and then 5 more family members in random order (because the 3rd brother doesn't have to sit next to his siblings, right?)

So, we have 4 possible arrangements of the first 4 siblings: 2x2x1x1=4

And the five other family members (brother + 4 grandparents) can sit in 5!=120 possible arrangements.

Overall, all possible arrangements are therefore (2)(2)(5!)=(4)(120)=480

So, why not B?
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 29 Jul 2014, 14:01
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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.

Kudos points given to correct solutions.

Try NEW Combinations DS question.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 30 Sep 2014, 16:55
Bunuel why did you do (6-1)!? We have 6 units, so couldn't they be arranged in 6! ways?
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 30 Sep 2014, 21:30
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p2bhokie wrote:
Bunuel why did you do (6-1)!? We have 6 units, so couldn't they be arranged in 6! ways?



This is because it is a case of "CIRCULAR PERMUTATION"

Suppose if 5 Boys, A, B,C, D and E are standing in a straight line and we have an arrangement ABCDE. Now, if we move each boy one place to the right then arrangement will become EABCD. ABCDE and EABCD are not the same. So we calculate N! or 5! arrangements when dealing with straight line arrangements.

In the above case, we are dealing with Circular arrangements. Now suppose the same 5 Boys are seated around a circular table, then we have an arrangement ABCDE. If we move each boy one place to the right the arrangement will still be ABCDE. So we are counting the same arrangement more than once and this would be wrong. So we subtract the duplicates from N! and hence we get (N-1) ! arrangements.

Hope this helps !!!
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 01 Oct 2014, 05:37
Thanks Bunuel. The explanation definitely helps. Thanks again.

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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 30 Mar 2015, 01:58
2 sisters => 2!.
sisters between any 2 brothers => 1) arrangement of 2 brothers out of 3 where order matters => 3P2 = 3!.
leaving aside 2 S and 2B(chosen one's), we have 1B and 4GPs and one combined unit of BssB => 6 units to be arranged.
Circular => 5!.
ans = pdt of all = 2! * 3! * 5! = 120 * 12 = 1440.
Ans C.
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 28 Sep 2015, 13:13
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I solved it by drawing a little illustration:

Attachment:
Unbenannt.jpg
Unbenannt.jpg [ 25.31 KiB | Viewed 3966 times ]


There is one restriction, so that the two sister should be seated between two brothers. So first ignore the two sisters and calculate the circular permutation of n = 7 (9-2 sisters).

circular permutation of 7 = (7-1)! = 6! = 720

Now the two sisters can sit in 2! ways

Hence, total combinations: 2*720 = 1440
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 04 Oct 2015, 00:31
Can any one tell what is wron in my aproach.

Total ways of seating 9 people in a circle is 8!

And ways in which the sisters will not be inbetween brothers is when all three brothers are together. I.e. (6-1) 5!.

Now subtracting, 8! - 5! should't give the ways in which 2 sisters are seated bw brothers..?
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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New post 04 Oct 2015, 00:52
goldfinchmonster wrote:
Can any one tell what is wron in my aproach.

Total ways of seating 9 people in a circle is 8!

And ways in which the sisters will not be inbetween brothers is when all three brothers are together. I.e. (6-1) 5!.

Now subtracting, 8! - 5! should't give the ways in which 2 sisters are seated bw brothers..?


With your approach you are dealing with big numbers which is imo not very efficient. Mind that the wys in which the sisters do not sit together is not complete when all brothers sit together. You can have e.g. the following combination: BRO SIS BRO BRO SIS GP GP GP GP --> Now does the Brothers sit together? No! Are the two sisters in between them? No!

Try to visualise and for this example look at the graph I did above, it may help you?
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Re: Nine family members: 5 grandchildren (3 brothers and 2 siste  [#permalink]

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