What is the logic behind the following permutations Given four letters $a, b, c, d$.
If $b$ has to come after $a$ and $d$ has to come after $c$, how many ways it can be done.
The solution is: $\frac{4!}{(2! * 2!)}$.
But what is the reason behind it?
Thanks
 A: The conditions given effectively reduce this to a problem of arranging $xxyy$, where the $x$ positions will then be replaced with $ab$, in order, and likewise with $y$ and $cd$. Arrangements of letters with repeats is simply the factorial of the total count divided by the factorial of each repeat count, $\frac {4!}{2!2!}$ as expected.
A: The reason why it the answer is $\frac{4!}{2\cdot 2}$ can be explained by: The "$a$ before $b$" and "$c$ before $d$" are two independent events, so we can straightaway divide by two twice.  
You can also look at it this way:
As there are four (equally likely) possibilities that can occur when we arrange $a,b,c,d$: 


*

*$a$ before $b$, $c$ before $d$ (favourable)

*$a$ before $b$, $c$ after  $d$ (not favourable)

*$a$ after  $b$, $c$ after  $d$ (not favourable)

*$a$ after  $b$, $c$ before $d$ (not favourable)


however only one out of the four equally likely possibilities is favourable, and we have a total number of $4!$ cases, thus the answer if $\frac{4!}{4}=6$. Hope it helps.
