Circular Arrangements

[akhilsuhag]

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?

[GMATGuruNY]

For CIRCULAR arrangements:

1. Place one element in the circle.
2. Count the number of ways to arrange the REMAINING elements.

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?

Strategy:
Since the 2 sisters must sit in adjacent seats, first place one of the 2 sisters in the circle.
Then count the number of ways to arrange the REMAINING people.

After one of the sisters has been placed in the circle:
Number of options for the other sister = 2. (To the left or right of the first sister.)
Number of options for the seat to the left of the 2 sisters = 3. (Any of the 3 brothers.)
Number of options for the seat to the right of the 2 sisters = 2. (Either of the 2 remaining brothers.)
Number of ways to arrange the remaining 5 people = 5!.
To combine these options, we multiply:
2*3*2*5! = 1440.

[MartyMurray]

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?

You can work your way out starting with the two sisters.

The two sisters go together as a unit, and they can be arranged 2 ways.

So that's 2 arrangements.

Next to the two sisters are two brothers. The two brothers outside the two sisters can be arranged 2 ways.

We have 2 ways to arrange the sisters and 2 ways to arrange the two brothers who surround the sisters.

So, so far we have 2 * 2 = 4 arrangements.

If we are choosing from three brothers two to surround the sisters and one left over, there are 3 ways to choose 2 from 3.

So we have 2 ways to arrange the two sisters, 2 ways to arrange the two sister surrounding brothers and 3 ways to choose the two brothers.

So far 2 * 2 * 3 = 12 arrangements.

At this point, each of the 12 BSSB arrangements can be considered one element, (BSSB).

For each of those 12 (BSSB) arrangements, there are one brother, B, and four grandparents, G G G G, left to arrange with it.

So we get a total of 6 elements, (BSSB) B G G G G.

Arrangement of 6 elements around a table is 5! = 120.

So for each (BSSB) B G G G G we get 120 arrangements around the table.

12 different BSSB arrangements * 120 different (BSSB) B G G G G arrangements = 1440 Total Arrangements

[Matt@VeritasPrep]

Mitch's explanation is a great combinatorial approach, but let me supplement that with a fun probabilistic one.

Suppose we had NO restrictions at all. There are (n - 1)! ways to arrange n people around a circular table, so we'd have 8! ways to arrange our nine people.

Now let's figure out what fraction of those arrangements meet our conditions. Let's seat the first sister first. If we randomly seat the second sister, there is a 2/8 chance that she'll sit next to the first sister. (There are 8 seats at the table, 2 of which are next to the first sister.)

Now let's deal with the grandparents. If they DON'T take the seats next to the sisters, then we'll be left with the brothers forced to take those seats. When the first grandparent sits down, there's a (5/7) chance (s)he doesn't sit next to the sisters. (There are 7 seats left, 5 of which are not next to the sisters.) The next grandparent would be 4/6 (since there are only 6 seats left after we seat the first grandparent), the next would be 3/5, and the last would be 2/4.

The probability of all these events = (2/8) * (5/7) * (4/6) * (3/5) * (2/4) = 1/28. So 1/28 of our arrangements will have the two sisters together with two brothers on either side. Hence our answer is 8! * (1/28), or 1440.