Linear Arrangement with 2 constraints for placement

Question

In how many ways can a teacher in a kindergarten school arrange a group of 6 children (Susan, Karen, lei, Tim, joy and Zen) on 6 identical chairs in a straight line so that Susan is on the left of Tim and to the right of Joy?

we've seen questions before like this where you'd find 6!/2 , but what happens when we add another factor here like another constraint? how do we generalize it so that we can think through these more complex constraints

Thank you

chetan2u · Answer

You can look at it in two ways. 
1) 6 can be arranged in 6! ways but three of them can be in ONLY one arrangement whereas we have considered 3! for these 3 persons. 
So answer =6!/3!=6*5*4=120
2) Choose 3 seats in 6C3 ways, now these three can be seated in just one way in each of these 6C3 ways. But the remaining 3 can occupy the remaining seats in 3! ways. 
6C3*3!=6!/3!=120.

Lazybum · Answer

Aren't there more than 1 arrangement for the 3 individuals? They do not have to sit directly next to eachother so long as Susan is in between Tim and Joy.
e.g.
K,L, T,S,J, Z
 T KL S Z J

so isn't that more than 1 type of arrangement?

In a similar question where there are 6 people with names A,B,C,D,E and F. A must be behind B. We use 6!/2. It doesn't seem like we can use a similar rule here.

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edit: your second explanation clarifies it very well. I just understood.

in those 3 seats, you have only 1 way to arrange these 3 people since Susan must always be in the middle of the other two. This is a very nice combination of using combinations and counting principals.

thank you,

Linear Arrangement with 2 constraints for placement

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