Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6

Question

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. How many such arrangements are possible ?

A. 360
B. 120
C. 80
D. 240
E. 60

Bunuel · Accepted Answer

Total # of arrangement of 6 people is 6!. In half of the cases Susan will be seated left to Tim and in half of the cases Susan will be seated right to Tim (why should one seating arrangement have more ways to occur than another?). So, # of arrangements to satisfy the given condition is 6!/2=360. Answer: A. Similar questions to practice: in-how-many-different-ways-can-the-letters-a-a-b-91460.html six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html Hope it helps.

ind23 · Answer

Hi,

When you are using the above method what you are assuming is that they are sitting next to each other always, which is not what the question states.
The question only says that S is always sitting left of T, maybe next, maybe away 1 chair ...there is no constraint on that.

-----------------------------------
 Kudos, if the post helped

GyanOne · Answer

Total number of arrangements = 6! = 720

In exactly half, Susan will be to the left of Tim, which gives us 360 arrangements

Option (A)

satsymbol · Answer

Why Can not I use Glue method here? 
SK together, with 4 others - 5! = 120 ways.

KarishmaB · Answer

This post discusses the symmetry concept: https://anaprep.com/combinatorics-linea ... -symmetry/

Kinshook · Answer

Given: Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6 identical chairs in straight line so that Susan is seated always left to Tim. 
Asked: How many such arrangements are possible ?

Total arrangements possible = 6! = 720
In half of the arrangements Susan is seated left to Tim, number of arrangements so that Susan is seated always left to Tim = 720/2 = 360

IMO A

aditsatt · Answer

Hi Bunuel If are taken as 1, wouldn't the possible arrangements just become 5! = 120? Same logic as another question you posted about 7 children ABCDEFG where CF have to be seated together: There the answer was 6! *2 (since CF & FC are possible, which is not the case here with ST) - https://gmatclub.com/forum/seven-childr ... l#p3619794

Bunuel · Answer

In that other question, S and F had to be seated  immediately  next to each other, with no one in between. That’s why the total was the number of all possible arrangements (7!) minus the ones where they were together (6! * 2, because S–F and F–S are both possible), giving 7! - 6! * 2.

Here, it’s different. We’re told that Susan must always be seated  to the left of  Tim, not immediately next to him. That means Susan can be anywhere on Tim’s left side. So the total number of arrangements is 6! in all, and exactly half of those will have Susan left of Tim while the other half will have Tim left of Susan. So we just take 6!/2.

aditsatt · Answer

Gotcha, thanks  Bunuel !.

I feel I have a lot of issues with the extent to which I include cases (eg. Overseeing that S&T can be separated by others) when solving.

I will be very grateful for your tips on how I can be better at this.

Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6

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Susan, John, Daisy, Tim, Matt and Kim need to be seated in 6

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