Probability with numbered balls and bins I am trying to learn more about probability and came across an interesting question that I am stuck on and can no longer find online. There are 20 numbered balls and 10 bins. Someone is trying to assign the balls to the bins, but does it with replacement on accident.
So they did the following: Place a ball in bin 1, record it, then remove ball (with replacement remember). Place a ball in bin 2, record it, then remove ball. Place a ball in bin 3, record it, then remove ball. So for each bin, you have put in 1 ball. There are ten bins, therefore you do that process once for every bin. Once you have done that the experiment is over.
What is the probability exactly 1 ball was assigned to exactly 4 bins? What is the probability at least 2 bins received the same ball?
A) 1 Ball in 4 Bins:
We have ${20 \choose 1}$ being the different ways we can choose the 1 ball that was assigned. Also, we have ${19 \choose 6}$ being the different ways the other 19 balls can be picked for assignment. However, what is the sample size? Would it be $20^{10}$? Thus the answer would be $\frac{{20 \choose 1}{19 \choose 6}}{20^{10}}$.
B) Probability of at least 2 repeated can be represented as $1-P(\text{Zero Repeated})- P(\text{One Repeated})$. So $P(0) = {20 \choose 10}/20^{10}$ and $P(1) = \frac{{20 \choose 1}{19 \choose 9}}{20^{10}}$. Then we can plug and chug.
Are these right? Is this how to think about this type of problem?
 A: After clarification of the question it can be answered as following:


*

*What is the probability that exactly 1 ball was assigned to exactly 4 bins?
We have $\binom{20}{1}$ ways to choose the "4-fold" ball and $\binom{10}{4}$ ways to choose the bins where it should go. The other 6 bins can be arbitrarily filled with remaining 19 balls. Hence the overall number of combinations is:
$$
\binom{20}{1}\binom{10}{4}19^6.
$$
In this way we however double-count the cases where there are two balls each assigned to exactly 4 bins. There are $\binom{20}{2}$ such pairs and $\binom{10}4\binom{6}4$ ways to choose corresponding bins. The other 2 bins can be filled arbitrarily with remaining 18 balls. Bringing everything together the final result is:
$$
\frac{\binom{20}{1}\binom{10}{4}19^6-\binom{20}{2}\binom{10}{4}\binom{6}{4}18^2}{20^{10}}.
$$


*What is the probability at least 2 bins received the same ball?
The simplest way to answer this question is to use complementary probability  of the event "all bins receive different balls":
$$
1-\frac{\frac{20!}{10!}}{20^{10}}.
$$
Replacement of $\frac{20!}{10!}$ with $\binom{20}{10}$ would be wrong here, since after choosing $10$ balls out of $20$ there are still $10!$ ways to assign the balls to certain bins. 
A: The way I understand from the comments is that you are modeling this is by a function $f:\{\text{bins}\}\longrightarrow \{\text{balls}\}$ in which you take a bin and you assign a ball to it, they can have the same ball(replacement). so there are indeed $20^{10}$ possible functions. Now, for A, you pick the ball in $\binom{20}{1}=20$ ways and then you choose the other $6$ bins. But you are assuming those are going each to a different bin and you are not considering different orderings. This contradicts the supposition of replacement. This seems a little more involved that what you propose, because you need to know that you are not overcounting possibilities. I would use the Inclusion Exclusion Principle to compute $\left |\bigcup _{i=1}^{20}A_i\right |,$ where $A_i = \{\text{The i-th ball was assigned to 4 bins}\}.$ Notice that $|A_i|=\binom{10}{4}19^{6}.$ For 2 $|A_i\cap A_j|=\binom{10}{4}\binom{6}{4}(20-2)^{2}.$ Can you have $3$ balls going to $4$ bins? No, so you just have to combine this two possibilities. 
For B, You want to take out the functions that are one to one because if a function is not one to one then 2 bins were going to the same ball. So the probability would be $$1-\frac{\binom{20}{10}}{20^{10}}$$
