An enterprise has four departments. There are three managers in each

Question

An enterprise has four departments. There are three managers in each of the four departments. These managers are to be
seated along a circular table in such a way that all managers from the same department sit together. Two arrangements are
considered different only when the neighbours of any manager change. How many different arrangements of seating are
possible?

A.  144

B.  488

C.  6^5

D.  \frac{11!}{(3!)^4}

E.  11!

chetan2u · Accepted Answer

Hi..
Four departments and each has three managers, so managers in each can be seated in 3! Ways, so for all 4 depts, 3!*3!*3!*3!...
These four departments are arranged in circular table and can be arranged in (4-1)!=3!...

Total ways = 3!*3!*3!*3!*3!=6^5

C

ayshwar · Answer

Shouldn't the answer be (6^5)/2 as there is no difference between clockwise and anticlockwise seating?
Consider the similar case of a necklace.

rszarate · Answer

I have the same concern.  Bunuel , could you clarify please?
Thank you in advance.
R.

kk47 · Answer

In circular arrangements we divide by 2 only when anti-clockwise and clockwise arrangement does not matter. For e.g. in case of a necklace, imagine this : we can wear it from any direction; picture the necklace as a vertical circle facing towards you, since you can rotate this circle / necklace on its vertical axis and wear again, the clockwise and anti-clockwise arrangement becomes irrelevant. That is why we divide by 2 in case of a necklace.

For this sitting arrangement, clockwise and anti-clockwise are 2 different scenarios. Consider A, B and C around a circular table. Total arrangements are (3-1)! - 2! = 2. These 2 arrangements are nothing but clockwise and anti-clockwise versions of same arrangement. So how are they different you might think? Keep "A" as your reference point. In one arrangement B and C would be on right and left side of "A" respectively and in the other arrangement B and C would be on left and right side of "A" respectively. Hence, both these clockwise and anti-clockwise arrangements would be different.

thyagi · Answer

You are considering arranging 12 people around a circular table whereas the question asks us to arrange managers of 4 departments in a such way that no manager of one team takes a seat between managers of any other team..all the managers of one department must sit together.. one team has 3 managers and these 3 can be arranged in 3! Ways. Similarly for the other three teams.... Arranging 4 teams around a circular table ( 4-1)= 3!..

So first arranging managers of 4 teams = 3!^4
Arranging teams= 3!
Total arrangements = 3!*3!^4
6^5.

C

Posted from my mobile device

ayshwar · Answer

The question may seem basic at first glance. 4 departments that can be arranged in  3! ways  with 3 managers each, who can be arranged in  3!^4 ways , thus resulting in C seeming right.

However, dig this:
Say the departments (associated managers) are A (EFG), B (IJK), C (MNO) and D (QRS).
Reading every arrangement clockwise, for every arrangement
A (EFG) B (IJK) C (MNO) D (QRS)
there is an arrangement
A (GFE) D (SRQ) C (ONM) B (KJI)
that is supposedly different and thus also included in C, but is actually the same given no different neighbor for any of the managers.

Hence, the answer has to be  6^5/2 .

Note: This case is similar to that of a necklace in which every bead has the same neighboring beads no matter how it is put on.

Gemmie · Answer

Number of ways of arrangement of 4 departments in a circle: (4-1)! = 3! = 6

Number of ways of arrangement of 3 managers in each department: 3! = 6

=> Total number:  6 * 6^4 = 6^5

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An enterprise has four departments. There are three managers in each

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An enterprise has four departments. There are three managers in each

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