# In how many ways we can have seating arrangements? [closed]

The Prime Minister of XYZ calls a meeting of Chief Ministers of the states whose state boundaries touch ZZZ (there are total 4 such states) to discuss on the terrorism. In how many ways can they seat themselves at a round table if two of the four states Chief Ministers choose to sit together?

## closed as off-topic by Pragabhava, JMoravitz, projectilemotion, Namaste, JonMark PerryMar 11 '17 at 2:31

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• Welcome to Math.SE! The ZZZ ... is XYZ? So there are 4 states total or 4 states that have boundaries with XYZ? If everyone shifts one seat clockwise ... is that considered a different arrangement? – Bram28 Mar 8 '17 at 18:56
• Thanks @Bram28 ZZZ is country (say Pakistan) and XYZ (is the country India) that has 4 states (Punjab, Gujrat, J&K and Rajasthan) which touch the international border with ZZZ (Pakistan). Each state has its own Chief Minister ( so total 4) out of which two wants to sit together (in any arrangement, say in the middle both are sitting together, or at the corner and so forth). – Madhu Sareen Mar 8 '17 at 19:02
• So ... do you have 4 people sitting around the table .. or 5? That is, does the Prime minister from XYZ sit at the table as well? That is still not clear to me. ... also, if the table is round, there is no corner :) – Bram28 Mar 8 '17 at 19:04
• Sorry.. there are 5 people including PM. and yes there are no corner. :) – Madhu Sareen Mar 8 '17 at 19:07
• The unnecessary storyline is unnecessary and potentially adding to the confusion. Regardless, these "circular seating arrangement" questions all rely on the same trick: have an arbitrarily selected person sit at the table first. (However you decide to select them is fine so long as there is only one choice. E.g. pick the youngest person, the shortest person, the smallest number, etc...). Now, describe any arrangement of people at the table in terms of that person's perspective. Choose who sits to the right of him, choose who sits to the right of that person, etc... until al seats filled – JMoravitz Mar 8 '17 at 19:08

To summarize what I said in the comments above:

Approach by multiplication principle. Begin by arbitrarily assigning labels to the people to be seated calling them $A,B,C,D,E$ where $A$ and $B$ are the ones who wish to sit together.

• Step 0: Seat $A$ at the table in an arbitrarily selected position. Important: This step does not contribute to our multiplication principle because making different choices here does not actually lead to different outcomes

• Step 1: Since $B$ wants to sit next to $A$, choose whether $B$ is sitting to the right of $A$ or to the left. ($2$ options)

• Step 2: Moving clockwise around the table from where $A$ and $B$ are currently sitting in some order, fill the next available empty seat with one of the three remaining people. Choose which person fills that seat. ($3$ options)

• Step 3: Continuing clockwise, pick who occupies the next empty seat. ($2$ options)

• Step 4: Continuing clockwise, pick who occupies the final empty seat ($1$ option)

Apply multiplication principle and reach a final conclusion.

Let's say we have 5 people A,B,C,D,E that sit at this table, and that A and B want to sit next to each other.

I'll assume that 1) everyone shifting one seat clockwise does not count as a different arrangement (though everyone sits in a different chair when we do so, everyone will still have the same person to their left and the same person to their right).

I'll also assume that 2) a mirror arrangement does count as a different arrangement (though everyone will still have the same two people they sit next to, the person that sat on their left will not on their right, and vice versa).

With this, let's start with seating A and B. Because of 1), it doesn't matter which seat I put A, but because of 2), B can be on the left of A or on the right, so there are 2 possibilities for that.

Once A and B are seated, let's put a third person clockwise from them, for which there are 3 possibilities, and for the last two people we then have 2 options left.

So: we get 2 (arranging A and B) * 3 (seat 3rd person) * 2 (seat last two people) = 12 possible arrangements