A Counting Problem Suppose letters $ a, b, c , d$ are to be arranged such that $a$ should always come before $b$ and so with $c$ should  always come first before $d$.  By brute force, I got the following arrangements:
$$abcd,acbd,acdb,cdab,cadb,cabd$$.
Now, there are six ways in arranging $a,b,c,d$. My question now is, what is the combinatorial way of finding the number $6$? Thanks.  
 A: One way to do it is this: There are $4!=24$ arrangements of all four letters with no restrictions. In half of those arrangements, $a$ precedes $b$, and in half of them, $c$ precedes $d$. The order of $a$ and $b$ and the order of $c$ and $d$ are independent, so we multiply $\frac12\cdot\frac12=\frac14$. One fourth of $24$ is $6$.

Here's another way: place $a$ and $b$ first, and there's only one way to do that. Now we get to place $c$ and $d$, and they get to go in the three spaces surrounding our $a$ and $b$ (before $a$, between the two, and after $b$). We know which order they go in. There are $\binom31=3$ ways to put them adjacent, and $\binom32=3$ ways to put them non-adjacent. That makes six.
A: Exactly half of all arrangements have $a$ before $b$, and exactly half have $c$ before $d$.  Hence your answer is $$\frac{4!}{2\cdot 2}=\frac{24}{4}=6$$
A: There are $4$ chairs in a row. There are $4!$ ways to arrange our $4$ people in a row.
There are $\binom{4}{2}$ ways to choose the pair of chairs to be occupied by $a$ and $b$. Once we have done that, where everybody sits is determined.
Thus the required probability is $\frac{\binom{4}{2}}{4!}$.
