Combinatorics with a condition. Jane is giving gifts to 3 sets of  cousins who are brother-sister pairs. 
She gives the gifts one after the other to her 6 cousins on the condition  that no brother receives a gift before his sister.  
In how many ways can this be done? 
I just need an approach for it as it is not of the type that I have done before.. .. 
 A: Lets call the brothers and sisters $a,b,c,A,B,C$ respectively (where $a$ is brother to $A$, etc...)
Arrange the sisters in any order.  Suppose for illustration that it was order $A~B~C$.
Put the brother of the final sister appearing in line into the line somewhere behind his sister.  $A~B~C~c$
Put the brother of the second to last sister appearing in line into the line somewhere behind his sister (need not be adjacent).

 There are three allowable positions for $b$ pictured here $A~B~\underline{~}~C~\underline{~}~c~\underline{~}$

Put the remaining brother into the line somewhere behind his sister (need not be adjacent).

 Suppose for illustration that the result of the previous selection was $A~B~C~b~c$.  Then for this step, the available choices are pictured as $A~\underline{~}~B~\underline{~}~C~\underline{~}~b~\underline{~}~c~\underline{~}~$

Count how many choices are available for each step and verify that having made this sequence of choices creates a good arrangement exactly once as well as that every arrangement is achievable by such a sequence of choices.  Conclude using multiplication principle.
A: We use the notation of JMoravitz. Suppose the $6$ people are arranged at random, with all arrangements equally likely. Then there are $6!$ arrangements. We count the favourables, where each brother is (somewhere) to the right of his sister. The argument uses symmetry. It will be phrased probabilistically, but it can be reframed using counting. 
The probability a is somewhere to the right of A is $\frac{1}{2}$. Given that a is to the right of A, the probability b is to the right of B is $\frac{1}{2}$. And given both of these have happened, the probability c is to the right of C is $\frac{1}{2}$. So the probability every brother is somewhere to the right of his sister is $\left(\frac{1}{2}\right)^3$, and therefore the number of favourables is $\frac{6!}{2^3}$. 
