Five persons are seated at a round table. Find the number of ar-
rangements such that all do not have the same neighbours in any
two arrangements.
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Answers
Answer:
There are five people and five seats. Hence, there are 5 ways to seat the first person, 4 ways to seat the second, 3 ways to seat the third, 2 ways to seat the fourth person and the firth person goes to the last remaining seat, so over there are 5! ways to seat five people.
Counting rotation equivalent seatings: What if we consider some of the seatings equivalent and the table is a round-table. For example, rotating people along the table clockwise or counterclockwise does not change their mutual position. Then, there are five possible actions that leave seatings equivalent, denote them by {I, R1, R2,R3, R4} . I is for an identity, that leaves everything unotuched, R1 moves every person clockwise to the next seat i.e. 1→2→3→4→5 and 5→1 , R2 moves every person clockwise by two sets i.e. 1→3→5→2→4 and 4 →1 . In a similar manner we can construct the arrangements R3 and R4 . You can also think of R4 as R−11 i.e. every person is shifted by one seat counterclockwise and R3=R−12 every person is shifted counterclockwise by two seats. Usually, while denoting actions we drop the arrows and employ the cycle notation as follows
IR1R2R3R4=(1 2 3 4 5),=(2 3 4 5 1),=(3 5 2 4 1),=(4 2 5 3 1),=(5 4 3 2 1).
These actions form a cyclic group C5={I,R1,R2,R3,R4}, a group that acts on seatings via rotation. The question is, how many inequivalent seatings we have now? And the answer is
#C5(5)=1|C5|(5!+0+0+0+0)=4!(1)
This makes perfect sense, since we partition the total set of 5! seatings into equivalent arrangements called orbits. Why we used the term partition here? It is because, (a) each set (orbit) of 5 arrangements does not overlap with other orbits (i.e. by applying the group action to any seating we can reach overall only 5 seatings) and (b) these arrangements exhaust all 5! possible seatings. Note that in (1) we have five terms (5!+0+0+0+0) , each of this term represents the number of fixed points, another important term. In practice, it is much easier to count the size of fixed point sets rather than the orbits, nevertheless, the number of orbits is what we are after. A fixed point is an operation in C5 or some other group G that leaves seatings unchanged. In our example, all four rotations R1, R2,R3, R4 change the seatings, expect the identity I. Hence, we have 0 of fixed points for the rotations and 5! fixed points for the identity I, this is because regardless of the arrangement we choose, the identity I will leave it unchanged. In this example as you can see, counting fixed points is easy, in the example that follows it will be a bit more tricker, but still much easier than counts orbits directly. Now, the question is, how are the orbits and the sets of fixed points related? The answer is given by the Cauchy-Frobenius-Burnside lemma, that states
|O|=1|G|∑π∈G|fixG(π)|(2)
and this is exactly how we calculated (1), in our case G=C5.
Counting rotation and reflection equivalent seatings: What if we extend the cyclic group by including the reflections along symmetry axes? Then, the set of possible actions becomes D5={I,R1,R2,R3,R4,F1,F2,F3,F4,F5} , where Fk are the flip motions along axis k . Since a pentagon has five axes of symmetry, there are exactly five additional flip motions. This group is called a dihedral group, which is the group of symmetries of a regular polygon, and in our case, it is a pentagon, that is why D5. The figure below demonstrates all 10 actions.
Using the cycle notation, the reflection actions are
F1F2F3F4F5=(1)(2 5)(3 4),=(2)(1 3)(4 5),=(3)(2 4)(1 5),=(4)(1 2)(3 5),=(5)(1 4)(2 3).
For example, in F1=(1)(2 5)(3 4), here (1) denotes the vertex (seat) that is not altered by F1 and (2 5) denotes the operation of interchanging people seating at seats with numbers 2 and 5 , similarly, people at seats 4 and 5 swap their seats. So how many seatings are possible under the rotations and reflection symmetries? The answer is
#D5(5)=1|D5|(5!+0+0+0+0+5⋅2+5⋅2+5⋅2+5⋅2+5⋅2)=120+5010=17.
The fixed points of the rotational motions have been discussed earlier, except I whose fixed point set’s size is 5! , there are no other fixed points. For the flip motions, we have a point (vertex) that lie exactly on the symmetry axis (there are five of them), seating any of 5 people on that seat, and applying a corresponding flip motion, will not change the person’s position. Next, for the remaining 4 people we have 2 groups of 2 people, these two groups of people interchange the seats as a whole, the two people seating in each of the groups do not change the seats, this contributes to the fixed point set and hence the size of the fixed point set is 5⋅2 .
In general, a regular polygon with n sides has 2n different symmetries: n rotational symmetries and n reflection symmetries. So, for a table with n people we have
#Dn(n)=1|Dn|(n!+the size of fixed point set under all reflection motions).
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