An experiment consists of drawing 10 cards from an ordinary 52-card pack. If the drawing is done with replacement, find the probability that no two cards will have the same face
value.
The solution to this almost trivial if we work with ordered samples. Indeed, there are 13 ways to select the first kind, and 4 ways to select a suit for that kind. Next, for the second card there are 12 kinds left and again 4 ways so select a suit. There are
$$(13\cdot 4)(12\cdot 4)\cdots(4\cdot 4) = 52\cdot48\cdots 20\cdot 16$$
ways to generating hands of 10 cards in which the order counts and the kids are all different. The fact that that sampling is with replacement has no importance at this point since we are not allowed to have duplications in terms of face values anyway. For the denominator, we note that there are obiously
$$52^{10}$$ ways to generate hands of 10 cards where the order counts with replacement. So the required probability is
$$
p = \frac{52\cdot48\cdots 20\cdot 16}{52^{10}}.
$$
See also here for a similar problem.
Now the question: How would you go if you were forced to reason in terms of unordered samples? At present, I can only see that the denominator must be
$$
{52+10-1\choose 10},
$$
but I don't see a strategy for correctly computing the numerator…
Best Answer
The answer you found of $\dfrac{52\cdot 48\cdot 44\cdots 16}{52^{10}}$ is correct. It could have been written more simply by not talking about all $52$ cards but rather talking about the $13$ ranks and ignoring suit. It could also have been written more cleanly by using the notation for falling factorials that $n\frac{k}{~} = \underbrace{n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}_{k~\text{terms}}$ letting you write the answer as:
$$\dfrac{13\frac{10}{~}}{13^{10}}$$
As for unordered sampling... the punchline is going to be that whether you actually intentionally remember the order in which the sampling occurred or not it is more convenient to have pretended that you had kept track for the purposes of calculation. The outcomes counted by stars and bars are not equally likely outcomes and so the denominator should not involve the calculation for $\binom{52+10-1}{10}$ regardless whether or not there exists a choice of sample space which could have been described as having that many elements.
Remember that the probability of an event $A$ from a sample space $S$ will equal $\Pr(A)=\dfrac{|A|}{|S|}$ when the outcomes in $S$ are known ahead of time to be equally likely events. In general however you may not calculate $\Pr(A)$ in this fashion. There are two outcomes to playing a lottery. You win or you lose. You do not win the lottery with probability $\dfrac{1}{2}$ however.