Q:

combinatorics - poker

I have an assignment in school about combinatorics. 
Information given: three ppl are playing poker. There are three open cards on the table and two cards in each persons hand. On the table we'll find an ace, one queen and one 10. One person, let's call her Kate, has a queen and 
a 10. 
There's a lot of questions and scenarios but there's one specific case I have a hard time grasping. So here it goes: If the other two ppl are to have a straight, they both need to pick up a king and a jack. How many different possible ways can they both pic up a straight. Knowing there's still four cards left of both kings and jacks this seems easy. 
$${4\choose1} \times {4\choose1} \times {3\choose1} \times {3\choose1} = 144$$ 
However, this answers seems to be right if we count the two hands as another set of cards if they would swap hands. I want to calculate the amount different possible ways if the opposite (them switching cards) counts as the same, (Hope this makes sense).
I guess I could just divide 144 by 2 which equals 72. But it doesn't make sense to me. I want a better understanding of this for the following scenarios etc. 

A:

You should treat is as two events    
$\frac {\binom{4}{1} \binom{4}{1}}  {\binom{47}{2}}$  
$\frac {\binom{3}{1} \binom{3}{1}}  {\binom{45}{2}}$

