+1439 At a meeting, four scientists, two mathematicians, and a journalist are to be seated around a circular table. How many different arrangements are possible if the mathematicians must sit next to each other? (Two seatings are considered equivalent if one seating can be obtained from rotating the other.)
+189 The problem fails to state the number of available seats at this table, but I will assume for problem-solving purposes that there are 7 seats at this circular table.
There are 7 guests in total, but both mathematicians must sit beside each other. Because of this, we should consider the group of 2 mathematicians as 1 entity that occupies 2 seats. The remaining 5 non-mathematician entities can freely occupy the remaining 5 seats. This effectively reduces the problem to arranging 6 entities around a circular table.
6 entities have 6! seating arrangements. However, note that table is circular. If all the entities rotated one seat to the left, then the relative seating positions would remain the same. The entities could rotate a total of 6 times before returning to their original seat, so we have effectively overcounted by a factor of 6. Account for this by dividing by 6.
Also, the mathematicians are in a group of 2, so there are 2 ways to arrange the mathematicians within its group.
Therefore, the number of arrangements is \(\frac{6!}{6} * 2 = \frac{720}{6} * 2 = 120 * 2 = 240\).
