Doubt about why I can't treat this as a Bernoulli process I know the title is not descriptive enough, but I don't know how else to say it. I don't know why I can't use the Binomial distribution to get the result I'm looking for. The teacher solved it long ago with a more straightforward method, but I tried today to solve the same problem using Binomial distribution and the result is not the same.
So, the problem is:
A total of $17$ balls are distributed in $20$ baskets, where every basket can hold as many balls as I want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishsable from another. What is the probability of every ball falling into the first basket?
The teacher solved it through combinatorics, using:
$$\dfrac1{\dbinom{20+17-1}{17}}$$
I wanted to solve it using Binomial distribution:
$$\dbinom{n}k p^k (1-p)^{n-k} = \dbinom{17}{17} \left(\dfrac1{20}\right)^{17} \left(\dfrac{19}{20} \right)^0$$
Knowing that there are $17$ throws, and I want $17$ successes. And as there are $20$ baskets, and only one is the successful one, the probability of each ball of going into that basket is $1/20$.
The problem is, it doesn't give the same result. And I don't know why, because I think I'm not using the binomial distribution wrong.
Thank you for your time!
 A: The key is that the balls are indistinguishable.
Let us consider an example, where we can work out explicitly. Consider $2$ balls and $3$ baskets.
If the balls are distinguishable, say colored red and blue, then the first ball can go into any of the three baskets, while the second ball can also go into any of the three baskets. This thereby gives us a total of $3 \times 3$ options.
$$\{\color{blue}1,\color{red}1\},\{\color{blue}1,\color{red}2\},\{\color{blue}1,\color{red}3\},\{\color{blue}2,\color{red}1\},\{\color{blue}2,\color{red}2\},\{\color{blue}2,\color{red}3\},\{\color{blue}3,\color{red}1\},\{\color{blue}3,\color{red}2\},\{\color{blue}3,\color{red}3\} \tag{$\star$}$$
where $\{\color{blue}i,\color{red}j\}$ means blue ball chose basket $i$ and the red ball chose basket $j$. Hence, the probability that both balls fall in the first basket is $\color{green}{\dfrac19}$.
If the balls are non distinguishable, then let us list out the options.
$$\{2,0,0\},\{0,2,0\},\{0,0,2\},\{1,1,0\},\{0,1,1\},\{1,0,1\}$$
where $\{m_1,m_2,m_3\}$ means the baskets $i$ has $m_i$ balls. Essentially we are counting $(\star)$, but we now consider $\{\color{blue}1,\color{red}2\}$ and $\{\color{blue}2,\color{red}1\}$ to be the same since there is no color to distinguish between them. Hence, the probability that both balls fall in the first basket is the same as the probability that the first basket contains both the balls, which in this case is $\color{green}{\dfrac16}$.
Hence, in general, if we want to distribute $m$ balls in $n$ (distinguishable) baskets, the total number of ways is


*

*$n^m$, if the balls are distinguishable.

*$\dbinom{n+m-1}{n-1}$, if the balls are indistinguishable.


Clearly, both intuitively and algebraically, we have $\dbinom{n+m-1}{n-1} < n^m$.
Hence, in general, the probability that all the $m$ balls fall in the first basket is


*

*$\dfrac1{n^m}$, if the balls are distinguishable.

*$\dfrac1{\dbinom{n+m-1}{n-1}}$, if the balls are indistinguishable.

A: I think you may have some confusion (as do we) about what you mean by the first basket.
If you mean basket number 1 (or 7 or 12 or any other particular number) then your formula will give the correct result $\left(\frac{1}{20}\right)^{17}$.
If you mean the basket that the first ball drops into then you are interested only that the remaining 16 balls fall into that basket $\left(\frac{1}{20}\right)^{16}$.
A: Your answer $\left(\frac{1}{20}\right)^{17}$ makes sense to me, given your description of the problem. 
I recognize the pattern of the teacher's answer, and that is an answer to a different question. There are $\binom{20+17-1}{17}$ ways to distribute $17$ indistinguishable balls into 20 urns. If all such distributions were equally likely, then that would explain the teacher's answer. However as you describe the problem, not all such outcomes are equally likely. As you describe it, $$16,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ is $17$ times more likely to occur than $$17,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ but your teacher's answer counts these outcomes as equally likely. Perhaps that was an error or perhaps that was somehow stated in the original wording.
