Mike and fred are at a conference and have been broken into a group with 10 others people. There is 2 tables that are circular, where one seats 7 and the orher has exactly enough to set the rest. How many different ways can the group be seated if mike and fred must sit together??
To solve this problem, we can consider Mike and Fred as a single entity. This means that we have 11 entities (MikeFred, and the other 10 people) to be seated around 2 circular tables.
Now let's break down the problem further:
First, let's consider the table that seats 7 people. Since the table has exactly enough space for the rest of the people (10 - 2 = 8 people), we need to find the number of ways to arrange these 8 people around the table.
The number of ways to arrange these 8 people around a circular table is (n-1)!, where n is the number of people to be seated. In this case, (8-1)! = 7! = 5040.
Now, since Mike and Fred must sit together, we can think of them as a single entity (MikeFred). So now we have 10 entities (MikeFred and the other 8 people) to be seated around the larger circular table with enough space for everyone (10 people).
The number of ways to arrange these 10 people around the circular table is (n-1)!, where n is the number of people to be seated. In this case, (10-1)! = 9! = 362,880.
Finally, since we have two different tables, we need to account for the different ways to assign people to each table. In this case, since the tables are identical, the assignment is not unique. We divide the total number of arrangements by 2.
Therefore, the total number of different ways the group can be seated if Mike and Fred must sit together is:
5040 * (362,880 / 2) = 13,476,480.
So, there are 13,476,480 different ways the group can be seated if Mike and Fred must sit together.