Q:

Probability of guessing 3 numbers out of 9 (Order matters)

My friend is playing a game and he needs to solve a puzzle which we are arguing about.
The puzzle is picking 3 numbers out of a group of 9.  Each time you pick a number, it is removed from the pool so there are no duplicates.
I came up with there being 504 possibilities (9*8*7).  I also wanted to figure out the odds of getting the right combo, so I did 1/9 + 1/8 + 1/7 = 191/504.
Am I right so far?
I also wanted to find the percent which would be 191/504 = ~37%?
I think I'm wrong, can someone confirm?

A:

Assuming the order of selection does not matter, there are $\binom{9}{3}=84$ possible ways to pick three numbers from the nine, and only one of which is the correct selection.
The probability then of getting the correct selection is $\dfrac{1}{\binom{9}{3}}=\frac{1}{84}\approx 0.0119$
Note, this can also be found using $\frac{3}{9}\cdot \frac{2}{8}\cdot \frac{1}{7}$ which equals the same thing.  I.e. the probability that the first selection is correct times the probability that the second is correct given that the first is correct, times the probability that the third is correct given that the first two are correct.

If order of selection is important, then the probability is even lower.
There are $9\cdot 8\cdot 7$ different possible outcomes, only one of which is the correct choice.
The probability of having chosen the correct choice is then $\frac{1}{9\cdot 8\cdot 7}=\frac{1}{504}\approx .00198$
Again, this could be found using a probability argument instead of a counting argument.  The probability the first selection is correct is $\frac{1}{9}$.  Multiply this by the probability that the second selection is correct given that the first is, and again for the third selection, giving $\frac{1}{9}\cdot \frac{1}{8}\cdot \frac{1}{7}$ which again equals the same as above.

