Is it A?

i feel it is C

winnerhere wrote:Sixteen guests have to be seated around two circular tables, each accommodating 8 members. 3 particular guests desire to sit at one particular table and 4 others at the other table. The number of ways of arranging these guests is

1) 9c5

2) 9!(7!) / 4!5!

3) 9!((7!)^2) /4! 5!

4)(7!)^2

5) none of these

Tough question!

Let's break the task down into two parts:

1) assign the guests to the two tables; and
2) arrange the guests around the tables to which they've been assigned.

1) assigning the guests

We have 3 people at one table and 4 at the other. So, we need to assign the remaining 9 guests.

We can do so by focusing on either table (doesn't matter which one).

If we start with the 3 person table, we need 5 more to fill it up. There are 9 people available, so there will be 9C5 possible selections. Once we select those 5, the other 4 automatically sit at the other table.

If we start with the 4 person table, we need 4 more to fill it up. There are 9 people available, so there will be 9C4 possible selections. Once we select those 4, the other 5 automatically sit at the other table.

Why doesn't it matter with which table we start? Because 9C5 = 9C4!

(As an aside, n Choose k = n Choose (n-k).)

2) arranging the guests

When we arrange n objects in a line, there are n! ways to arrange them; when we arrange n objects in a circle, there are (n-1)! ways to arrange them.

Since there are 8 people at each table, there are (8-1)! = 7! possible arrangements for each table.

When we're making MULTIPLE selections/permutations, we always MULTIPLY the individual results. Accordingly, our final answer is:

9C5 * 7! * 7!

= 9!/4!5! * (7!)^2

= (9*(7!)^2) / (4!5!)

choose C!

Thanks Stuart.

When I encountered the problem - I apporoached the sum like this.

Among the 16 places 3 guests will go to a table 1 and 4 guests will go to a table 2. So remaining guests are 9.

For table 1 5 guests are needed. So 9C5 selections are possible. Reamining 4 guests will goto table 2.So total possibilities of selection is 9c5 x 1 = 9c5.

now we have 8 guests in each table. Each can be arranged in 7! ways. So (7!) x (7!) ways to arrange.

Finally 9C5 X 7! x 7!

I wanted to know if My approach is right
I guess its right way to approach this sum going by your answer