<s>[INST] If $n$ numbers are generated, what is the probability that the product of all those numbers is a multiple of 10? A computer generates random numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ (each has equal probability). If $n$ numbers are generated (with replacement), what is the probability that the product of all those numbers is a multiple of 10?

Attempt:
Let us see easier case, when $n=4$. The total number of possible random numbers can be counted as the number of non-negative solutions to 
$$ x_{1} + x_{2} + ... + x_{8} + x_{9} = 4$$
This is because, for example: $\{1,2,3,9\}$ can be seen as $x_{1}=x_{2}=x_{3}=x_{9}=1$ and all others are 0, then again $\{5,5,2,4\}$ as $x_{5}=2, x_{2}=x_{4}=1$ and the others 0. The number of solutions is $\binom{8+4}{4}$. 
Now we have to count the total number of possibilities when the product of the generated numbers is multiple of 10. If it is multiple 10, then it must contain a 5 and at least one from $\{2,4,6,8\}$. We can count by the following:


*

*2 generated numbers are $5$ and $2$, with no restriction for the other 2. (note that this can contain $4,6,8$). This will be the same as counting nonnegative solutions: $x_{1} + ... + x_{9} = 2$. Count is $\binom{8+2}{8}$

*2 generated numbers are $5$ and $4$, but this time there is restriction for the other 2: it cannot contain 2 because this was counted earlier. This is the number of nonegative solutions for: $x_{1} + x_{3} + ... + x_{9} = 2$. Count is $\binom{7+2}{7}$

*2 generated numbers are $5$ and $6$, but restriction for the other 2: cannot contain a 2 or a 4. Similarly we consider: $x_{1} + x_{3} + x_{5} + ... +x_{9} = 2$. Count is $\binom{6+2}{6}$

*2 generated numbers are $5$ and $8$, restriction for other 2: cannot contain a 2, 4, or 6. Similarly consider: $x_{1} + x_{3} + x_{5} + x_{7} + x_{8} + x_{9} = 2$. Count is $\binom{5+2}{5}$ 
So when $n=4$, the probability of interest is $\frac{\binom{8+2}{8} + \binom{7+2}{7} + \binom{6+2}{6} + \binom{5+2}{5}}{\binom{8+4}{8}}$
I want to double check my approach, and also looking for better answers.
 [/INST] You are assuming that getting $2,4,3,5$ is the exact same as getting $5,2,4,3$ and that somehow the probability of getting $\{2,3,4,5\}$ in any of the 24 orders is the exact same probability of getting $\{3,3,4,5\}$ any of its 12 orders, or getting $\{4,4,4,\}$ in its one order.
You are assuming that getting a set $\{2,3,4,5\}$ is equally likely as getting a set $\{5,5,5,5\}$.  And therefore you are counting (with great difficulty) the ways to get different sets of numbers.  
But each order is as equally likely as any other as there are $24$ ways to order $\{2,3,4,5\}$ but only $12$ ways to order $\{3,3,4,5\}$ there is no reason at all do do this.
Each $(a_1,a_2,a_3,a_4)$ tuple counts and each order is a seperate option.  
There are $9^4$ possible tuples.  
That's it.
So there are $9^n$ ways to pick $n$ numbers.
If you have at least one even and at least one $5$ you will have a multiple of $10$.
There are $4^n$ ways to have no evens or $5$s at all.  And there are $5^n$ ways to have no evens but maybe or maybe not have $5$s.  ANd there are $8^n$ ways to have no $5$s but maybe or maybe not have evens.
By inclusion exclusion:
Number of ways to not have a even and a 5=  Number of ways to not have an even + Number of ways to not have a 5 - (due to double counting) Number of ways to not have both=
$5^n + 8^n - 4^n$.
So probability of not being a multiple of $10$ is $\frac {5^n + 8^n - 4^n}{9^n}$
So probability of being a multiple of $10$ is $1 - \frac {5^n + 8^n - 4^n}{9^n}$
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