We define and enumerate circular permutations.
In a circular permutation, all positions on the circle are considered equivalent. Thus,
the position of an object is solely determined by its position relative to the other
objects. By contrast, the objects in ordinary permutations have absolute positions-
first, second, third etc. An ordinary permutation can be thought of as a linear
- First, we select the objects to be placed in the circular permutation.
This can be done ways.
Second, we arrange the objects in a circle and use the FPC. When the first object is placed in the circle, all of the positions are equivalent, so there is only 1 choice. Once the first object is placed, the remaining positions in the circle become distinct. Hence, there are choices for the position of the second object in the circle, choices for the third object, and so on. Thus, the number of ways to arrange the objects in a circle is . Finally, we multiply the number of ways to select the objects by the number of ways to arrange them in a circle to obtain To see that this is equal to , we note that
- Applying the proposition with yields that the number of such permutations is
Here, we are counting the number of circular permutations of objects (taken at a time). According to the corollary, the number of ways for the 10 children to march in a circle is .
According to the corollary, the total number of ways for them to be seated around the table with no restrictions is
Next, we count the number of seating arrangements that have Merlin and Galahad next to each other. We seat them first, in 2 possible ways- with Galahad on Merlin’s left and with Galahad on Merlin’s right. Then the other 3 members can be seated in 3! = 6 ways. Thus there are ways for the group to be seated in violation of the restriction.
Finally, by subtracting (using the Rule for Complements), we conclude that the number of ways to seat the group so that Merlin and Galahad are separated is .
The first child, say a girl is seated in 1 way, since all seats at a round table are considered identical. Once she is seated, there are 3 seats left for the other 3 girls and therefore they can be seated in ways. The 4 boys can then be seated in the remaining 4 seats in ways.
The total number of ways to seat the eight children is
First, we have to choose the 10 children to be seated. This can be done in ways. Next, we can seat the first child in 2 ways (horse or tiger) and then the other 9 children can be seated in ways relative to the first child.
Hence, the total number of configurations of 10 of the 20 children for the first carousel ride of the day is
At first, this may seem like a straight forward circular permutation, but unlike the other scenarios we have seen, the round object in this case can be flipped upside down to create another (different) circular permutation.
Thus, we count the circular permutations of objects and then divide by 2, to get different bracelets can be created.