From Discrete Mathematics by R Johnsonbaugh, problem 6.5.24.
"In the California Daily 3 game, a contestant must select three numbers among 0 to 9. One type of “box play” win requires that three numbers match in any order those randomly drawn by a lottery representative, repetitions allowed. What is the probability of choosing the winning numbers, assuming that the contestant chooses three distinct numbers?"
The answer says $\frac{3!}{10^3}$. I understand why the denominator is $10^3$ but why is the numerator $3!$ and not say $10\times9\times8$?
And if I accept the numerator $3!$ above, why does the example below not have $6!$ in the numerator? After all, in the example below, just like the one above, we match n different numbers in any order.
problem 6.5.25: "In the Maryland Lotto game, to win the grand prize the contestant must match six distinct numbers, in any order, among the numbers 1 through 49 randomly drawn by a lottery representative. What is the probability of choosing the winning numbers? " Answer: $\frac{1}{C(49,6)}$.