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In a race, 15 runners are numbered from 1 to 15. Find the probability that: 4 of the first 6 finishers have a single digit number.

We are considering the order here, so there are $15^{(6)}$ total possible ways to order any possible way. There are $9^{(4)}$ ways for single digit. Then there are $6^{(2)}$ for a double digit for 2 of them.

But this gives me, $P = 18/715$ when the actual answer is something else?

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    It is easier to work with combinations in my opinion. There are $\binom{15}{6}$ ways to choose the first six finishers (where order within those chosen doesn't matter). There are $\binom{9}{4}\binom{6}{2}$ ways to pick exactly four singledigit racers and two two-digit racers. – 2017-01-20
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    @JMoravitz, why doesnt order matter in this case? – 2017-01-20
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    Because each possible ordering of the first six is equally likely and each case where order doesn't matter that we care about is equally represented amongst those possibilities where order does matter. – 2017-01-20

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Probability = $\frac{\binom{9}{4} \cdot \binom{6}{2}}{\binom{15}{6}}$

As using combination we does not need to take care of arrangements.

Hope you get correct answer on solving.

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$15^{(6)}$ counts the ways to arrange a list of 6 things selected from 15.   That is if you mean the falling factorial: $\frac{15!}{(15-6)!}$

$9^{(4)}$ counts the ways to arrange a list of 4 things selected from 9, and $6^{(2)}$ counts the ways to arrange a list of 2 things selected from 6.

Which is okay so far, but you have not counted ways to combine these two list into a single list of 6; by selecting two from places to put the double digits (in whatever order they come).

Thus probability you want is $\dfrac{\dbinom 62 9^{(4)}6^{(2)}}{15^{(6)}} ~=~ \dfrac{\dfrac{6!}{2!~4!}\dfrac{ 9!}{5!}\dfrac{6!}{4!}}{\dfrac{15!}{9!}} ~=~\dfrac{\dbinom{6}{2}\dbinom{9}{4}}{\dbinom{15}{6}}$

Which happens to be the probability of selecting two of six items and four of nine items, when selecting six from all fifteen items.