What is the Probability of Choosing 3 Specific Balls from a 15-Ball Set? My high school geometry textbook includes the following problem as an example:

Three pool balls are chosen at random from a set numbered from 1 to 15.  What is the probability that the pool balls chosen are numbered 5, 7, and 9?

My answer is $\frac{1}{455}$.  The text book's answer is $\frac{6}{455}$.  Which answer is correct?
I arrive at my answer, as follows: 


*

*The probability of the desired result is the ratio of the number of outcomes that yield the desired result to the total number of possible outcomes.  

*The desired result, balls 5, 7 and 9, together form a single 3-ball combination; a single outcome:
$$
1
$$  

*The total number of possible 3-ball combinations is:
$$
\frac{15!}{3!(15-3)!} = \frac{(15 \cdot 14 \cdot 13)}{6} = \frac{2730}{6} = 455
$$

*Thus, the ratio described in (1), above, is $1\colon455$, and the answer is:
$$
\frac{1}{455}
$$
There are at least two alternative approaches that yield the same answer:  


*

*Rather than using combinations, one might use permutations, in which case the numerator is the $3!$ permutations that yield the desired result, and the denominator  is the $\frac{15!}{(15-3)!}$ permutations that could occur in choosing 3 balls, which is equal to $\frac{6}{2730}$ = $\frac{1}{455}$.  

*Or, one could use the product of the odds of each selection being a success:  $\frac{3}{15} \cdot \frac{2}{14} \cdot \frac{1}{13}$ = $\frac{6}{2730}$ = $\frac{1}{455}$.


HOWEVER, my textbook says the solution is:
$$
\frac{3!}{\frac{15!}{3!(15 - 3)!}} = \frac{6}{455}
$$
The textbook's statement of the problem and solution is here:
Lesson 13-3, Permutations and Combinations: Problem 6.  I believe the textbook's solution compares a number of permutations in the numerator to a number of combinations in the denominator, which seems to me to be a matter of apples and oranges.

Your answer is correct.
 We have to consider either permutation or combination for both numerator and denominator  to get the correct probability.
