23 chips numbered 1 through 23 are placed into an urn. The chips are then selected one by one, without replacement, until all 23 have been selected. What is the probability that the 9th, 14th, and 21st chip so selected had numerical values that were less than or equal to five?
I think I should be using indicators to solve this, but I'm getting thrown off that I'm looking for 3 exact terms.  Any help would be appreciated.
 A: To count the favourable ways, you can separate the process in steps and then use the rule of product or multiplication principle:


*

*Choose $3$ numbers less than or equal to $5$. You can do that in $\dbinom{5}{3}=10$ ways. 

*Place these numbers in the $3$ given places ($9$th, $14$th and $21$st). You can do that in $3!$ ways. 

*Place the remaining $20$ numbers in the remaining places. You can do that in $20!$ ways.


Since there are (obviously) $23!$ equally probable ways to select the chips without replacement, the required probability is equal to $$\frac{\dbinom{5}{3}3!20!}{23!}=\frac{5!20!}{2!23!}=0.0056$$
A: As suggested in my comment above, you can reduce this question to that of "what is the probability that the first three picked are less than or equal to 5" by arguing that you could simply pick the spaces you are interested in first and then fill in the rest of the spaces later.
For the first chip, there are 5 numbers available out of 23 which are okay, so probability of the first being okay is $\frac{5}{23}$
For the second chip, there is one fewer 'okay' number, and one fewer number overall, so probability is $\frac{4}{22}$, and similarly the third probability is $\frac{3}{21}$.
As the events are independent, you can multiply the probabilities together:
$\dfrac{5\cdot4\cdot 3}{23\cdot 22\cdot 21} = 0.0056$ agreeing with Stef's answer above.
