Mathematicians' Dinner Party

Since Andy- Bandy & Bandy-Candy sit together, this means Andy, Bandy & Candy all sit together. Fixing one place out of 6 for Bandy means selecting one out of six identical seats, therefore only one way. Now, the adjacent seats to Bandy are taken by Andy & Candy, in two ways ( Andy to Bandy's right or to his left). Remaining 3 places can be taken by 3 people in 3! ways. By the principle of multiplication, the complete job is done in 3! x 2 x 1 ways, therefore 12 is the correct answer.

First of all, since Andy and Bandy must sit beside one another; and Bandy and Candy must sit beside one another, it follows that Bandy is seated between Andy and Candy. We can count them altogether as one group. This leaves us with 4 'groups' to arrange. By the formula of circular permutation ( $p(n)=(n-1)!$ ), we get that there are 6 ways to arrange them. However, there are two ways to seat Andy and Candy. Hence, $2\times 6=12$ . There are 12 ways to seat them.

Since Andy and Bandy want to sit next to each other, so do Bandy and Candy, so we give Bandy a constant sit, then Andy and Candy can sit by 2 sides of Bandy. There are 3 sits left for Dandy, Endy and Fandy so we can find 6 ways for 3 of them to sit. By using symmetry when Andy and Candy have an exchanging for their sits, We can find 6 other ways which makes the total ways are 12. ( Because rotations are counted as the same seating arrangements so there is no more ways).

Andy, Bandy, Candy will sit together and Bandy will be in between Andy and Candy.So they can be seated in following two different ways. Andy Bandy Candy Candy Bandy Andy

As the above three will sit together they can be considered as a single person.So there are 4 person who will sit around the circular table.Now n person can sit a circular table in (n-1)! ways. So,total number of arrangements 2*(4-1)!=12

The total number of arrangements for n objects around a circle is (n-1)! , because, rotations are counted as single arrangement.

In the problem A and B want to sit together and also, B and C want to sit together. Therefore, we treat 'ABC' as a single object and treat D,E and F as distinct objects. Hence, we get 4 objects in all to arrange around the circle.

Thus, the number of permutations are (4-1)!=3!

But, A and C can exchange their positions inside the collective object ABC in 2! ways.

Hence, total permutations are 3! x 2! = 12

Let's call all the candidates A, B, C, D, E & F

Given constraints: 1) A & B sit together; 2) B & C sit together

Above two constraints can be satisfied if B sits between A & C i.e. in following manner:

                    A B C   or    C B A

So there are two ways to arrange A, B & C, to satisfy the constraints.

Lets put A-B-C in a block and call this block as "Z"

Now our problem boils down to finding out the no. of ways for Z, D, E & F to sit in a circular table.

Concept: Total no. of ways for 'n' people to sit in a circular table = (n-1)!

How did we get the above expression? It is the direct consequence of the fact that rotations give the same sitting arrangements. This is explained below:

We have 4 blocks to be arranged in circular table i.e Z, D, E & F.

So, we are fixing one seat as the head seat (as a marker) to keep track of rotations.

Z can be placed on any of the four seats( recall, block Z contains 3 seats) Therefore, there are 4 ways to place Z.

Next, 3 seats are left. So, D can be placed at any of the 3 seats left. Hence, there are 3 ways to place D.

Similarly, there are 2 ways to place E.

Only last seat is left for F. So, there is only one way to place F.

This gives us: 4 * 3 * 2 * 1 = 4!

4! is the no. of ways to seat 4 people in a straight line as well.

But our solution is not yet complete.

Let's place a pole at the center of the circular table and rotate it clockwise with Z, D, E & F sitting on one seat each. Z is placed at the head seat and head seat has moved to the position where the seat next to it was positioned before rotation and all other seats also replace the seats placed next to them.

What do we observe?

We observe that the positions of Z, D, E & F have changed on table but relative to each other, they are still sitting in the same arrangement. So, this arrangement is the same as before rotation.

As there are 4 seats, it will take 4 rotations for Z (head seat) to return to its original position. After 4 rotations, all the four seats will return to their original positions. So, if we had 'n' seats and we had to seat 'n' people in a circular table, it will take 'n' rotations for seats to return to their original positions.
We already understand that during these 'n' rotations, relative positions of people sitting will NOT change. They are all the same arrangements.

So, we will divide n! by n, for the reason given above, i.e

No. of ways = n! / n = n * (n-1)! / n = (n-1)!

For 4 blocks, as in this case,

No. of ways = 4!/4 = 3! =6

Now, we had block Z which contains 3 seats, seating A, B & C. We had calculated no. of ways as 2, such that constraints are satisfied.

Finally,

Total no. of ways to seat A, B........., F, such that given constraints are satisfied:
= 3! * 2 =12

Correct ans. = 12

Draw a diagram of the table, labelled A-F clockwise. Circle A, B, and C. Draw a line between A and B, and B and C, realising that A, B, and C must be next to each other, with B in the middle, in order for A to speak to B, but for B to also speak to C. We can therefore ascertain that either ABC, or CBA form half the table, with D, E, and F able to sit in any of the three other spaces in any order. So, we can either make a list of ABC: FED; FDE: EDF; EFD; DEF; and DFE, seated clockwise around the table, and also ACB: FED; FDE; EDF; EFD; DFE; and DEF as the only other possibilities. We can now count up our list, to make 12 various seating arrangements.

Yet by using permutations: As there are 6 friends, we would expect to form 6! = 720, which is 6x5x4x3x2x1 = 720. However, we can see from our diagram that ABC, three of the friends can only be sat in two ways, A, B, C, or C, B, A, as B (Bandy) must be sat between A (Andy) and C (Candy), in order to talk to both of them. Therefore, the permutation needed for all the possibilities of their order here is 2! = 2, or 2x1 = 2. However, the other three friends, FED, can be seated in any order in the three remaining seats, creating 3! = 6, which is 3 x 2 x 1 = 6. In order to therefore calculate the total permutations for all 6 friends, we need to multiply 2, by 6, to make 12.

use (4-1)!*2

Since Andy and Bandy need to sit next to each other, and Bandy and Candy need to sit next to each other, Bandy has to be sitting between Andy and Candy i.e. the three of them have to be fixed either in the position ABC or CBA.

Given that they are sitting at a circular table, for convenience's sake we will put them at the top of the circle, and seat everyone else in clockwise order i.e. (ABC / CBA) _ _ _ because rotations mean that ultimately if you start counting from ABC / CBA in clockwise order you will count for three more people.

To sort D, E, and F into the three slots requires 3!, and for each of the 3! ways they can either be sitting with ABC or CBA (i.e. 2 ways).

By Multiplication Principle, 2 * 3! = 12 = Your Answer.

andy,bandy,candy must sit together so we group them together... andy and candy can interchange the seat ... so we get 2! dandy,endy, fandy can sit freely ... so we get 3! so ans shd b 2!X3! = 12

Since Andy, Bandy, Candy must be seated next to each other in a circle, let A, B, C be one group. There are a total of (4-1)! ways to permute ABC, D, E and F in a circle. Among Andy, Bandy, Candy, there are 2 ways to arrange them, either ABC or CBA. Thus, total number of ways is (4-1)! x 2!

Since Bandy is between Andy and Candy, we can treat this group of three as just one object. So we only have four objects to permute around a circle. This can be done in $(n-1)! = (4-1)! = 3! = 6$ ways. But the group of three can be arranged clockwise (or counterclockwise, it doesn't matter) as Andy-Bandy-Candy or Candy-Bandy-Andy. Hence, there are a total of $6*2 = 12$ ways to seat the six friends.

First, group A,B,C together, since they must sit together, then there will be 4 "groups", including the 3 individuals D E F, Hence there are 3! to group the 4 groups. However, A must sit next to B, and B must sit next to C. Therefore there are 2 ways to group A B C. Therefore there is a total of 3!*2=12 ways.

Let the guys be named A,B,C,D,E,F now consider the ring (ABC)(D)(E)(F) and treat ABC as one person. The four persons can be permuted 3! times to obtain different arrangements and according to given condition there can be two arrangement for (ABC) - ABC , CBA So by multiplication principle we get 3! x 2 = 12

We denote 6 friends with A,B,C,D,E,F. From the restriction A,B,C are always seat in one group i.e. ABC or CBA. Then we can assume there only 4 persons on table so we have 3!=6 posibilities Finally we have 6 x 2=12 ways to arrange them

Since B sits next to both A and C, the three have to sit either as ABC or CBA. The other three slots can be filled by D, E and F in 3! (6) ways. Since it's 6 for ABC and 6 for CBA, the total is 12.

Only two ways to satisfy AB and BC next to each other. They are ABC and CBA. Each of the above has 6 combinations with D, E, and F. Therefore the total is: 2 x 6 = 12

for circular permutation of n objects the value is (n-1)!/2 in this problem, A & B, C&D,E, F are to be taken as 4 distinct people to be placed, because A&B always sit together, and similarly C&D always sit together, however A&B and C&D can be arranged in themselves in 2! ways, so the required value is (4-1)! x 2 x 2/ 2 =12

6 friends ABCDEF are present to be seated on 6 seats. A(Andy), B(Bandy) and C(Candy) want to sit together so they will treated as one person. So we have 4 combinations to seat them on 6 seats. Calculations: Hence No. of arrangements= 1/2 (4!) = 12