Probability of a runner explanation 
In a race, the 15 runners are randomly assigned numbers $1 \to 15$ inclusive. Find the probability that: B The fifth runner to finish is the 3rd finisher with a single digit number.

The runners are obviously distinguishable So let $S$ denote single and $D$ denote double digit.
We require the sequence: $DSSD\textbf{S}$,where the last $S$ is constant (cannot change) and the 4 before it can be rearranged. 
Here is my work: Can you spot a mistake OR a mistake in my explanation?
(1) We have $\binom{4}{2}$ ways of arranging $DSSD$ sequence.
(2) We have $\binom{9}{2}$ of picking a single digit number for these middle $S$.
(3) We have $\binom{6}{2}$ for double digit in $D$
(4) We have $\binom{7}{1}$ ways for choosing a single digit for the last $S$.
(5) We have $\binom{5}{2} \cdot  \binom{15}{5}$ ways of ordering the $DSSDS$ and choosing any $5$ numbers. 
Question:
Our selection of $2$ numbers for $S$, $\binom{9}{2}$ say picks two numbers $(1, 2)$, but this excludes $(2, 1)$ so say I had $D_1S_1S_2D_2$, that means $(S_1, S_2) = (1, 2)$, but what if I want $(S_2, S_1) = (1, 2)$? Since that is also a possibility isn't it? How does this take care of it?
 A: The first four can be any four element subset of $15$, all with equal probability. There are ${9\choose 2}\cdot{6\choose 2}$ such subsets containing exactly two Ss and two Ds.  Conditioning on this having happened there are $7$ Ss and 4 Ds left. The probability that the fifth runner is an S is then ${7\over11}$. The probability $p$ we are after therefore is given by 
$$p={{9\choose 2}\cdot{6\choose 2}\over{15\choose4}}\cdot{7\over11}={36\over143}\ .$$
A: You go off the rails in item 5. Once you've picked which of the runners are in the top five (which you've counted correctly), the number of rearrangements of the first four finishers is $4!$ and the number of rearrangements of the last ten is $10!,$ so just multiply what you have by those two. I think it's that simple.
A: I am giving a method you may find simpler.
It uses your first point, and then just multiplies probabilities, thus
$\dbinom42\times\left[\dfrac6{15}\cdot\dfrac9{14}\cdot\dfrac8{13}\cdot\dfrac5{12}\right]\times\dfrac7{11}= \boxed{\dfrac{36}{143}}$
