Partial Order Relations and Total Order Relations Question:
A={1,2,3}
1) How many partial order relations can be induced over A ? 
2) How many total order relations can be induced over A ? 
3) Does A exist a transitive relation? 
I guess total order relations over A is P(3,3)=3!=6, but I don't know how to count partial order relations over A. 
Any help will be greatly appreciated!

[UPDATE]
According to Scott's reply, I get the following result,
Type 1: 3!=6
Type 2: 3
Type 3: 3
Type 4: 3
Type 5: 3!=6
So there are 21(6+3+3+3+6) partial orders on A.
Is it right? I hope someone can check it.thanks.
[UPDATE2]
Thanks for Henry's help. 
Type 4:6 
Type 5:1 
So there are 19(6+3+3+6+1) partial orders on A.
I am not sure I am right but I hope to get corrected.
3)Does A exist a transitive relation? 
I saw the answer(odd numbered problems have answers) is {<1,2>,<2,1>,<1,1>,<2,2>,<3,3>}.
However, I don't understand why <3,3> is included in the relation.
 A: Counting the partial order relations is a bit messy. Perhaps the most straightforward way is to organize them by their ‘shapes’. In the following diagrams the order is from bottom to top.


*

*They can be linear:
 *  
 |  
 *  
 |  
 *


*They have a minimum element but no maximum:
 *   *  
  \ /  
   *  


*They can have a maximum but no minimum:
   *  
  / \  
 *   *


*They can have two related elements and one unrelated element:
 *  
 |   *  
 *


*They can have three unrelated elements:
 *   *   *

You’ve already counted the partial orders that are linear: there are indeed $3!=6$ of them. Now you just have to count the partial orders of types (2)-(5) above. To get you started, in type (2) any one of the three elements of $A$ can be the minimum element. Once you’ve chosen that, however, the partial order is completely determined:
    1   2                  2   1  
     \ /        and         \ /  
      3                      3  

are the same partial order, just drawn differently. Thus, there are $3$ partial orders of type (2) on $A$.
