M30-02

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

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.

Another way -

I call it imagine people entering a meeting room one by one/ VeritasKarishma's method.
Refer-https://gmatclub.com/forum/combinatorics-made-easy-206266.html
Topic name-circular arrangements

our answer is ways in which 2 sisters(S1,S2) can sit between 2 brothers(B1,B2)+ arrangement of 1 brother(B3) and 4 grandparents (G1,G2,G3,G4)

Consider an empty circular table and sister S1 comes in-

She can sit in (1 way)

Next,the second sister S2 has 2 options one to the left or one to the right of sister S1 (2ways)

Next,we can select 2 brothers among three in 3c2 ways ie 3 ways who are B1 and B2.
B1 can sit in 2 ways (either to the left or to the right of two sisters) and B2 can sit in 1 way.

Now there are 5 people remaining B3,G1,G2,G3,G4 and 5 positions so they can sit in 5! ways.

Final answer-
1*2*3*2*1*5! =1440

Solve using a circular table diagram(keep space for 9 people) it will make more sense.

Hi,

I have approached this problem slightly differently.

1. We chose 1 brother in 3C1 ways and place him on a circular table in 1 way
2. We chose 1 sister in 2C1 ways and place her on the table in 2 ways. (left or right of the brother)
3. The second sister can sit on the table in 1 way only
4. We chose 1 brother in 2C1 ways and place him on the table in 1 way
5. Remaining 5 people can sit in 5! ways

My answer: 3*1*2*2*1*2*1*5! = 2880

VeritasKarishma Bunuel: Could you please tell me whats wrong with my method?

There is double counting in your method. Say you have S1, S2 and B1, B2, B3.
Say, for your first brother, you select B1 and make him sit anywhere. Then you chose S1 on his right and S2 on further right. Then you select B2 to sit on S2's right. And the others are placed around.
Now imagine for your first brother, you selected B2. Then you chose S2 on his left, S1 on her left and you chose B1 to sit on S1's left. Then all the rest around.
In your answer, these are two different cases but in the actual answer, this is a single arrangement only.

Simply put, say you want to arrange 5 people such that A and B sit next to each other on a circular table, you must consider them as a unit and arrange them in 2 ways.
If you instead use your method, you will select one of them in 2 ways and put the other next to him in 2 ways (left or right) doubling the arrangements.

Hence, whenever you need to make some people sit together, it is always better to use the method of tying them together as one unit and then arranging.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I did by using only one mathematical formula (the only one I remember from high school because I am just starting to review the quant content).

All of us know that (N-1)! applies for circular arrangements. N is the number of individual elements.

In this case, we only have 6 units in total despite the fact there are 9 people. The reason why is that the two sisters between the two brothers work as a single unit. If not, we would not be meeting the problem statement.

Thus, we have (6-1)! that is 120. However, within the group of four the sisters can change chairs so that we have 120 x 2 = 240.Furthermore, the brothers can also change chairs. We have 3 of them; therefore, this is 3x2 different arrangements for the brothers (why 3x2? because "XYZ", X can be A, B or C, so that we have 3 options, and YZ can be the left two and be commuted).

Consequently, we have 240x3x2=1440.

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Bunuel

May i ask why you choose to Consider treating the two brothers and the two sisters as one single unit, represented as {BSSB}.

That you choose to treat the sisters as 1 unit, i get that and we use the glue methode, because we are calculating the number of times the sisters are sitting together.

But we donot have to count the number of times that the brothers are sitting together. Besides, you go even a step further by treating 2 of the brothers together with the 2 sisters as one unit. To be honest i donot follow your reasoning behind your solution. Can you please elaborate a little bit more on this?


Thank in advance!

­
First of all, in the {BSSB} configuration, the brothers are not sitting directly next to each other. Next, we consider {BSSB} as a single unit because we need the sisters not only to sit next to each other but also to be positioned immediately between two of the brothers, as specified in the problem. This arrangement ensures that the requirements are met precisely, hence the formation of the {BSSB} unit.­