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25 Jul 2014, 09:09
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C
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Difficulty:
95%
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Question Stats:
49% (02:33) correct
51%
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wrong
based on 1046
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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 immediately between some pair of brothers.?
A. 120
B. 480
C. 1440
D. 2880
E. 8640
A. 120
B. 480
C. 1440
D. 2880
E. 8640
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25 Jul 2014, 09:10
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SOLUTION
In how many ways can nine family members, consisting of four grandparents and five grandchildren (three brothers and two sisters), be arranged around a circular table so that the two sisters sit next to each other and are positioned immediately between two of the brothers?
A. 120
B. 480
C. 1440
D. 2880
E. 8640
Consider treating two brothers and two sisters as a single unit, represented as {BSSB}.
With this grouping, we have a total of 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}, where {G} represents a grandparent and {B} represents a brother outside the unit.
These 6 units can be arranged around a circular table in \((6-1)!=5!\) ways.
Next, we'll analyze the {BSSB} unit in more detail:
We can choose 2 brothers out of the 3 for the unit in \(C^2_3=3\) ways.
Within the unit, these selected brothers can be arranged in 2! ways, such as \(\{B_1, S, S, B_2\}\) or \(\{B_2, S, S, B_1\}\).
Similarly, the sisters within the unit can be arranged in 2! ways, either \(\{B, S_1, S_2, B\}\) or \(\{B, S_2, S_1, B\}\).
Therefore, the final answer, considering all the possible arrangements, is given by \(5!*3*2*2 = 1440\).
Answer: C
Try NEW Combinations DS question.
In how many ways can nine family members, consisting of four grandparents and five grandchildren (three brothers and two sisters), be arranged around a circular table so that the two sisters sit next to each other and are positioned immediately between two of the brothers?
A. 120
B. 480
C. 1440
D. 2880
E. 8640
Consider treating two brothers and two sisters as a single unit, represented as {BSSB}.
With this grouping, we have a total of 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}, where {G} represents a grandparent and {B} represents a brother outside the unit.
These 6 units can be arranged around a circular table in \((6-1)!=5!\) ways.
Next, we'll analyze the {BSSB} unit in more detail:
We can choose 2 brothers out of the 3 for the unit in \(C^2_3=3\) ways.
Within the unit, these selected brothers can be arranged in 2! ways, such as \(\{B_1, S, S, B_2\}\) or \(\{B_2, S, S, B_1\}\).
Similarly, the sisters within the unit can be arranged in 2! ways, either \(\{B, S_1, S_2, B\}\) or \(\{B, S_2, S_1, B\}\).
Therefore, the final answer, considering all the possible arrangements, is given by \(5!*3*2*2 = 1440\).
Answer: C
Try NEW Combinations DS question.
25 Jul 2014, 18:17
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Bunuel
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 = \((6-1)! = 5!\)
\(2x3x2x5! = 12*120=1440\)
but there are also 3 seats left for three grandparents so 6720*3! *3! 
help 
