Urn model with replacement where each ball has a different number 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?
Since nobody in the comments could really help I came up with a different solution myself.

We can see this as an urn model with replacements and unordered samples. 

Now: 

$\Omega = \{ (k_0, k_1,\dots,k_9) \in \mathbb{Z}_+^{10} \colon \sum_{i = 0}^9 k_i = 4 \}$ 

and

$\mathbb{P} = \operatorname{Multin}( 10, (\frac{1}{10})_{c = 0}^{10})$.

Further, we have the event $A := \{ \omega \in \Omega_1 \colon \omega \text{ has 4 different digits} \}$.

We calculate:

$\mathbb{P}(A)\\  = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot |A| \\ = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot {10 \choose 4} \\ 
= {{4}\choose{(1,1,1,1)}}\cdot \frac{1}{10^4}  \cdot {10 \choose 4} \\ = 0.504$.

We can do this because ${{4}\choose{(1,1,1,1)}} = {{4}\choose{(1,1,1,1,0,0,0,0,0,0)}}$.

Actually, we could also come to the same solution using an urn model with replacement and ordered samples (using $\Omega = \{0,1,\dots,9\}^4$ with $\mathbb{P} = \operatorname{Uniform}(\Omega)$).
Since nobody in the comments could really help I came up with a different solution myself.

We can see this as an urn model with replacements and unordered samples. 

Now: 

$\Omega = \{ (k_0, k_1,\dots,k_9) \in \mathbb{Z}_+^{10} \colon \sum_{i = 0}^9 k_i = 4 \}$ 

and

$\mathbb{P} = \operatorname{Multin}( 10, (\frac{1}{10})_{c = 0}^{10})$.

Further, we have the event $A := \{ \omega \in \Omega_1 \colon \omega \text{ has 4 different digits} \}$.

We calculate:

$\mathbb{P}(A)\\  = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot |A| \\ = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot {10 \choose 4} \\ 
= {{4}\choose{(1,1,1,1)}}\cdot \frac{1}{10^4}  \cdot {10 \choose 4} \\ = 0.504$.

We can do this because ${{4}\choose{(1,1,1,1)}} = {{4}\choose{(1,1,1,1,0,0,0,0,0,0)}}$.

Actually, we could also come to the same solution using an urn model with replacement and ordered samples (using $\Omega = \{0,1,\dots,9\}^4$ with $\mathbb{P} = \operatorname{Uniform}(\Omega)$).