6 people are holding a show, one at a time, such that person $x$ has to go after person $y$ and person $z$. How many ways could the show be held? Let's say the people are called $a$, $b$, $c$, $x$, $y$, $z$
My initial thinking was to go by fixing "$x$" in a certain position, so:
$\underline {} \underline {} \underline {}\underline {}\underline {}\underline {x}$ 
Now for this configuration we have 5! combinations
$\underline {}\underline {}\underline {}\underline {}\underline {x}\underline {}$
For this one, after the $x$, only $a$, $b$ and $c$ can go, so that's $3 \cdot 4!$.
Similarly, I continued and got an answer in the $300$'s, which is not a possible given answer I have.
What's wrong with my method?
 A: I will assume that after is not restricted to immediately after. 
There are $6!$ ways that the people can be arranged without restriction. There are $3!$ relative orders for $x,y,z$. Exactly $2$ of these relative orders have $x$ after $y$ and $z$. So the number of ways is $6!\times \frac{2}{3!}$.
We can do the calculation more slowly. Imagine $6$ chairs in a row. We can choose $3$ of these to put 'Reserved" signs on in $\binom{6}{3}$ ways. We can then choose who of $x$, $y$, $z$ sits at which resereved chair in $2$ ways, for $x$ must sit on the last reserved chair. And then we can arrange the remaining people in the empty chairs in $3!$ ways, for a total of $\binom{6}{3}\times 2\times 3!$. 
A: We are given a poset $\{a,b,c,x,y,z\}$ where the only comparisons defined are $x>y$ and $x>z$.  We try to find how many ways there are to extend this to a chain.
First: either $x>y>z$ or $x>z>y$.  Two choices.  Without loss of generality, assume it is the first one.
Now, decide where to place $a$ in that list.  Either $a>x>y>z$ or $x>a>y>z$ or $x>y>a>z$ or $x>y>z>a$.  Four choices.
Now, decide where to place $b$ in that list.  Whichever was selected on the previous step, there will be five choices available.
Finally, decide where to place $c$ in that list.  Whichever was selected on the previous step, there will be six choices available.
The total number of ways then is:
$$2\cdot 4\cdot 5\cdot 6 = 240$$
A: You can make your method work.
If $x$ is in the last position, then there are $5!$ ways to arrange the other letters.
If $x$ is in the fifth position, then there are $3$ ways to select the letter in the last position from among $a, b, c$ and $4!$ ways to arrange the letters before $x$.  Therefore, there are $3 \cdot 4!$ arrangements with $x$ in the fourth position as you found.
If $x$ is in the fourth position, then there are $3$ ways to select the letter in fifth position and two ways to select the letter in the sixth position from among $a, b, c$.  There are $3!$ ways to arrange the letters before $x$.  Thus, there are $3 \cdot 2 \cdot 3! = 6 \cdot 3!$ arrangements in which $x$ is in the fourth position.
If $x$ is in the third position, then there are $2!$ ways to arrange $x$ and $y$ in the first two positions and $3!$ ways to arrange $a$, $b$, and $c$ in the last three positions, giving $2! \cdot 3!$ arrangements in which $x$ is in the third position.  
Since $y$ and $z$ must precede $x$, $x$ cannot be in the first two positions.  Hence, the number of arrangements in which $x$ follows both $y$ and $z$ is 
$$5! + 3 \cdot 4! + 6 \cdot 3! + 2! \cdot 3! = 240$$
That said, this method is less efficient than those of Andre Nicolas and JMoravitz.
Here is an alternate method:
There are six positions to fill.  We have six ways of placing the $a$, five ways of placing the $b$, and four ways of placing the $c$.  This leaves us with three positions to fill.  We must place the $x$ in the final open position.  There are $2!$ ways to place $y$ and $z$ in the remaining open positions.  Thus, the number of ways the show can be held so that $x$ appears after both $y$ and $z$ is 
$$P(6, 3) \cdot 1 \cdot 2! = 6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1 = 240$$
A: Simply represent as $abc\vee\vee\vee$ where each $\vee$ is a 'slot' where one of $x,y,z$ may be placed, although $x$ and $y$ must occupy the first $2$ slots.  So $a\vee bc\vee \vee$, $a\vee b\vee \vee c$, etc. possibilities. $a,b,c$ remain in same order.  Number of permutations is $P(6,3)={{6!}\over {3!}}$.  However the $\vee$'s cannot be distinguished so must divide by $3!$: $6!\over{3!3!}$ which is actually equal to $6\choose 3$.
Then must multiply by $3!$ for orderings of $a,b,c$ and $2!$ for orderings of $x,y$.
Final answer is $6\choose 3$$2!3!$.
