Understanding the $l!$ in the probability of $l$ events from a set of $N$ to happen at the same timestep out of $n$ total I know this is probably a simple question but honestly I've been struggling with it for a couple of days already.
I have $N$ independent events. Each of them can occur with equal probabilities in any of $n$ timesteps. I want to compute the probability of $l$ events to happen together. I find two expressions and I am not sure which one is the right one.
FIRST EXPRESSION:
The probability of the event $i$ happening in the time $t_j$ is $\frac{1}{n}$
The probability of the event $h$ happening in the time $t_j$ is $\frac{1}{n}$
...
The the proba of any event happening at time $t_j$ is $\sum^{N} \frac{1}{n} = \frac{N}{n}$  
Then, the probability of second event happening at $t_j$ would be  $\sum^{N-1} \frac{1}{n} = \frac{N-1}{n}$, because there are only $N-1$ events available. And then,
the probability of two events happening at time $t_{j}$ is $ \frac{N}{n} \frac{N-1}{n}$  
Following the same reasoning for $l$ events occurring together at time $t_{j}$:
the probability of $l$ events occurring at time $t_j$: $ \frac{N}{n} \frac{N-1}{n} \dots \frac{N-(l-1)}{n}$  
But I want to know the probability of events occurring at the same time, not only at $t_j$ but at any any $t_i$ in $n$,
the probability of $l$ events cooccurring at any time is: $ \frac{N}{n} \frac{N-1}{n} \dots \frac{N-(l-1)}{n} \times n = \frac{N!}{(N-l)! n^{l-1}}$
SECOND EXRPESSION:
The probability of the event $i$ happening in the time $t_j$ is $\frac{1}{n}$
The probability of the event $h$ happening in the time $t_j$ is $\frac{1}{n}$
...
Then, the probability of $l$ specific events to happen at time $t_{j}$ is $\frac{1}{n} \dots \frac{1}{n}$ ($l$ times).  
And since this could be for any of the $n$ timesteps,
the probability of this $l$ events to happen at any time is $\frac{1}{n^l}n = \frac{1}{n^{l-1}}$  
But here we considered only one specific set of $l$ events, to have all the possible ways to take $l$ from $N$ where the order is meaningless, we can compute the combination $\binom{N}{l}$. 
And we get: $\frac{1}{n^{l-1}} \binom{N}{l} $
QUESTION:  
We see that both expressions actually are "similar". They differ in a $\frac{1}{l!}$ factor. I guess this is usually a matter of considering the events undistinguishable or not... but honestly, not sure which one is correct in this case. 
Another not non-important issue, but that I can only solve after solving problem 1 is about normalization but I will ask that when i solve this one. 
 A: Your vision of the sum of probability is a bit distorted. Both the reasonings are not correct since you add possibilities of events that are not incompatible.
Here some examples:


*

*If you toss a normal dice, the probability of getting 1 is $1/6$, and also the prob of getting 2 is $1/6$. The probability to get 1 or 2 is the sum of prob $2/6$, since the events are incompatible: you can't get at the same time 1 and 2.

*If now you toss 2 dices, the probability to get 1 on the first dice is still $1/6$, and the probability to get a 2 on the second dice is also $1/6$. What is the probability of getting a 1 on the first dice or a 2 on the second dice? the possible couples of results are $36 = 6^2$, and the farovable results are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(3,2),(4,2),(5,2),(6,2)$ that are 11. the resulting probability is this $11/36$ that is different from the sum $2/6$. This happens because the two events can happen simultaneously when the result is $(1,2)$


In both your reasonings the probability that you're summing belongs to events that may happen at the same time!  


*

*In the first case, litterally the events happen at the same time (pun not intended). The real probability of any event happening at time $t_j$ is
$$1 - \left(\frac{n-1}{n}\right)^N$$
that is the opposite of the probability that no events take place at time $t_j$

*In the second case, if you take two distinct sets of $l$ events, they may happen at time $t_j$ together. In this case it is more difficult to point out the correct answer, since the events are not even independent.
The real answer is quite complicated and I don't know if there's a close form, sorry.
