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!
