How to know which is n and which is r ? When to use stirling number of second kind and stars and bars? From the first answer by Brian M. Scott in How many ways can 10 teachers be divided among 5 schools? I am getting a sense that there are 4 cases 
Objects          |   containers
distinguishable     | distinguishable 
indistinguishable   | distinguishable  -> stars and bars 
distinguishable     | indistinguishable  
indistinguishable   | indistinguishable 
My doubt is the second one .
If I say teachers are indistinguishable but schools are distinguishable and look at the question as "In how many ways 5 schools be allotted to 10 teachers " instead of seeing this as "In how many ways can ten teachers be divided among five schools?". Because I feel both mean the same.Then now Will I apply stars and bars and say answer is $\displaystyle\binom{10 + 5 -1}{9} $
But $\displaystyle\binom{10 + 5 -1}{9} \neq \displaystyle\binom{10 + 5 -1}{4}$
RHS is the correct answer posted by brian M scott in that question
LHS is my way when I exchange n and r due to confusion
In other words 
if $n =5 , r = 10 ,it\ is\ 14C4 $ <- Brians answer
if $n =10 , r = 5 , it\ is\ 14C9$ <- my confused interpretation lead to this
Question 1: so what is that factor which decides which is n and which is r ? Give me some examples to illustrate this 
Question 2: If this is the confusion , there are chances that I can even mix up  Stirling number of the second kind  and stars and bars concept usage .So please give me example to show when these two can be used .
 A: This answer probably adds nothing to the linked answer in the other question, but in case it helps:
One way to think of these problems is in terms of "urns and balls". Each ball goes into one  urn (this determines the mapping), analogy: each teacher is assigned to one school. 
Assume first that all balls are distinguishable (say, 5 balls: 1, 2 ... 5) and so the urns (say, 3 urns: A, B, C). 
Here are some possible assigments
   I         |     II         |    III        |    IV        |     V
 ------------ ---------------- --------------- -------------- --------------
A <- {1,3}   |  A <- {3,1}    |  A <- {1,3}   |  A <- {4,5}  | A <- {1,3,4} 
B <- {2,4,5} |  B <- {2,4,5}  |  B <- {2}     |  B <- {2}    | B <- {2}
C <- {}      |  C <- {}       |  C <- {4,5}   |  C <- {1,3}  | C <- {5}   

In all "urns and balls" scenarios (and in your four cases), there is no "order inside the urn". So, case II should never be counted as distinct from I - we'll always skip that.
Case 0: If we allow empty urns, (and if the balls and urns are distinguishable), all assignments are valid (I, III, IV, V ...), we have a total of $3^5=243$. You don't seem interested in this case.
Case 1:  If we disallow empty urns (each school must have at least one teacher), we must remove colum I from the count. Now (remembering that we have distinguishable balls and urns) the Stirling numbers of the second kind come into play: the total count is given by $3! \, S(5,3) = 6 \times 25 = 150$  (the factor $3!$ , permutations of the urns, takes into account that SN counts indistinguishable urns).
Case 2: If the urns are indistinguishable, (in the example, III and IV must be counted as one) we have instead the plain Stirling number of the second kind: $  S(5,3)=25 $ 
Notice that it's easy to go from urns indistinguishable to distinguishable (just multiply by the permutation of urns, $3!$). That's because, when urns turn distinguishable, each permutation maps to a new set of counts. That doesn't happen for the balls indistinguishable to distinguishable, because the order of balls does not matter (even if the balls are distinguishable) for balls that are inside the same urn, we must think it differently. And that's when stars-and-bars appears.
Case 3: If the balls are indistinguishable and the urns distinguishable what matters is how many balls go inside each urn (case III and IV are counted as one, again,
but for other motive). This is a fixed size (size=3) composition of the 5 balls, which by the  stars-and-bars argument, is ${5-1 \choose 3-1}=6$. 
Case 4: If the balls and urns are indistinguishable we have a fixed size (size=3) partition of the 5 balls. This doesn't have a closed formula, in this case the count is 2 (3-1-1 ; 2-2-1)
