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

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.
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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 = \((6-1)! = 5!\)

\(2x3x2x5! = 12*120=1440\)

I solved it by drawing a little illustration:



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

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 !!!

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?

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.

I love you sooooooooooooooo much Bunuel

I don't see where you've accounted for the restriction that the two sisters be between two brothers, and if you don't account for that, your answer will be much too big.

But the answer to the question, at a quick glance, I think should be 20,160, as written (or perhaps 9 times that, depending on whether we decide this is a circular permutation question). The solution above answers a different question than what is asked:



The phrase "so that 2 sisters are seated between any two of the three brothers", as written, means: if you pick any two brothers, you should find the two sisters somewhere between them. As the question is written, we should be counting an arrangement like this one as a valid arrangement:

BBGSGSGGB

for example, because if you pick any two brothers, then you'll find the two sisters between them somewhere (in some direction around the table). And if that's what the question means, in half of all possible arrangements you find the two sisters between any pair of brothers (because of the ten ways to arrange just the brothers and sisters, in half of them, SSBBB, BSSBB, BBSSB, BBBSS and SBBBS you satisfy the condition), which is how I got the answer above.

The question means to ask something very different from what it says; it means to ask how many arrangements are possible when the two sisters are seated *immediately* between *some* pair of brothers.

As written, the question is actually not even asking about circular permutations. The fact that things are in a circle does not, on its own, make a question into a circular permutation problem. It's only a circular permutation question if our concern is only about how things are arranged relative to each other, and if we're unconcerned about the specific seat each person is assigned.

I'll be happy to try to help, but I don't think I'm following how you're thinking about the problem. The product I highlighted in red, for example -- I'm not sure how you arrived at that.

I'd add that if you are planning to fix one person in place (and there's nothing wrong with doing that), it's not going to be easy if you choose to fix a brother. That will lead you to cases -- is it one of the brothers who is immediately next to the two sisters, or is it the other brother who isn't sitting next to his sisters at all? It's probably going to be easier to fix a sister in place, because you know that either sister has a sister to one side and a brother to the other side.

Bunuel why did you do (6-1)!? We have 6 units, so couldn't they be arranged in 6! ways?

Thanks Bunuel. The explanation definitely helps. Thanks again.

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bind the 5 kids together P2,2 * P3,3
arrange 4 grandparents with the bind p5,5

p2,2*p3,3*p5,5 = 1440

I got the correct answer of 1440, but I solved the question as if the people were all standing in a line (and not sitting in a circular arrangement.)

Why is the answer the same for both a line or circular arrangement? Shouldn't they be different, or is the question just trying to "trick" us by adding a complexity of having it be a circular arrangement?

Bunuel what about the seating arrangement of the grandparents? P = (4-1)! = 3! = 6
Shouldn't we multiply that by 1440 too?
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The arrangements of grandparents is included in 5! (check highlighted part in the solution above).
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Bunuel how about BSBSB or BSSBB or BBSSB

VeritasKarishma BrentGMATPrepNow IanStewart
as far as i know when it comes to circular arrangement i can start by fixing one person on one spot, so it is 1 way to do it that person is a grandparent

now about arrangement of brothers and sisters they can be arranged this way BBSSB


After fixing one grandparent on one spot, I`m left with 8 spots free, so for first brother 8 seats left, for second B, 7 seats left for first sister 6 seats for second S, 5 seats left and for third B 4 seats left. So i have 1*8*7*6*5*4 = 6720


but there can also be other arrangements such as BSBSB or BSSBB so three ways 3! 6720*3! wait I didnt finish yet but there are also 3 seats left for three grandparents so 6720*3! *3!

Whats wrong with my reasoning

The total ways of circular arrangement is n-1*1!however we cannot do it directly

first let us arrange the 2 sisters and 2 brothers in between =2*3*2(since there are 3 brothers and 2 sisters to chose from)

let us then consider the 4 people as 1 one group and arrange them with grandparenta and 1 remaning brothers since

there are 4 grandparent 1 brother and 1 group of brother and sister we can arrange them in circular arrangement in 5! ways

Since the group distribution has to be also taken into accountwe get
=5!*3*2*2
=>120*12
=>1440
Therefore C

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Thank you! Edited as suggested.
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