Suppose we have $N = 10$ balls where each ball has its own number. Thus, $C = \{0,1,2,3,4,5,6,7,8,9\}$. We draw $n = 4$ times and each time we put the ball back.
What is the probability to draw 4 balls where each ball has a different number?
The only solution I can come up with is:
If we draw a ball the first time there are $10$ allowed numbers. The next time we draw there are $9$ allowed numbers and so on. Thus, $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ draws that are allowed. There are $10^4 = 10000$ possible draws.
Now, we get $\frac {5040}{10000} = 0.504$.
Is there a different solution?