russ9 wrote:

Hi Karishma,

If there were constraints such as A can't be next to B or C, does that mean that we now have 5 seats but since 3 of them are fixed, the solution would be 2!/2?

I am assuming your question is this:

5 people are to be seated around a circular table such that A sits neither next to B nor next to C. How many arrangements are possible?

I don't know how you consider "...3 of them are fixed".

The way you handle this constraint would be this:

There are 5 vacant seats. Make A occupy 1 seat in 1 way (because all seats are same before anybody sits).

Now we have 4 unique vacant seats (unique with respect to A) and 4 people.

B and C cannot sit next to A so D and E occupy the seats right next to A on either side. This can be done in 2! ways: D A E or E A D

B and C occupy the two unique seats away from A. This can be done in 2! ways.

Total number of arrangements = 2! * 2! = 4

Check out these posts. First discusses theory of circular arrangements and next two discuss circular arrangements with various constraints:

http://www.veritasprep.com/blog/2011/10 ... angements/http://www.veritasprep.com/blog/2011/10 ... ts-part-i/http://www.veritasprep.com/blog/2011/11 ... 3-part-ii/ _________________

Karishma

Veritas Prep | GMAT Instructor

My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting

Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews