Does the order in which $N$ elements are distributed among $m$ groups affect the probability of achieving that distribution? For instance, let's consider the following problem: 

At a party, a clown is throwing $N$ sweets to $m$ kids, with $N,m>>1$
  and $N>>m$. If it is equally probable that a thrown sweet is caught by
  any kid (the clown is throwing the sweets in a perfectly random way),
  which would be the probability that the kid number 1 catches $N_1$,
  the kid number 2 catches $N_2$, and so on,
  $P(n_1=N_1,n_2=N_2,...,n_m=N_m)$?

My attempt
I have supposed that the first kids catches his $N_1$ candies first, then the second kid catches his $N_2$ candies, and so on, so that:
$$P(N_1,N_2,...,N_m)=P(N_1)·P(N_2|N_1)·P(N_3|N_1,N_2)·...·P(N_m|N_1,N_2,...,N_{m-1})$$
Then, each $N_i$ would follow a binomial distribution $\text{Bin}(p,n)$, being the available number of candies and the probability different in each case:
$$N_1 \sim \text{Bin} (p=\frac{1}{m},N) $$
$$N_2 \sim \text{Bin} (p=\frac{1}{m-1},N-N_1) $$
$$...$$
$$N_m \sim \text{Bin} (p=\frac{1}{m-(m-1)},N-N_1-N_2-...-N_{m-1}) $$

However, 


*

*Would it be my first supposition correct? That is, would they be all the possible orders in which the distribution $(n_1=N_1,n_2=N_2,...,n_m=N_m)$ can be achieved equally probable?

*Could (or should) the solution be approached in a different way?

*In general, does the order in which $N$ elements are distributed among $m$ groups affect the probability of achieving that distribution?
 A: Your notation is a bit confusing, but I think you mean that:
$$P(N_1) = {N \choose N_1} (\frac1m)^{N_1} ({m-1 \over m})^{N - N_1}$$
and 
$$P(N_2 \mid N_1) = {N - N_1 \choose N_2} ({1 \over m-1})^{N_2} ({m-2 \over m-1})^{N - N_1 - N_2}$$
The interesting thing is that when you multiply them you get a lot cancellation:
$${N \choose N_1} {N - N_1 \choose N_2} = {N! \over N_1! ~(N-N_1)!} {(N - N_1)! \over N_2! ~(N-N_1-N_2)!} \\= {N! \over N_1! ~N_2! ~(N-N_1-N_2)!} = {N \choose N_1, N_2, N-N_1-N_2}$$ 
where the last notation is a multinomial coefficient of dividing $N$ things into sets of sizes $N_1, N_2, N-N_1-N_2$.  Similarly the factor $(m-1)^{N-N_1}$ appears in the numerator of $P(N_1)$ and denominator of $P(N_2 \mid N_1)$.  So:
$$P(N_1) P(N_2 \mid N_1) = {N \choose N_1, N_2, N-N_1-N_2} (\frac1m)^{N_1+N_2} ({m-2\over m})^{N-N_1-N_2}$$
This continues through all $N_m$ in a similar pattern, and in fact we have:
$$P(N_1,N_2,\dots,N_m) = {N \choose N_1, N_2, \dots, N_m} (\frac1m)^N........(*)$$
So to answer your $3$ questions directly:


*

*Yes your chain reasoning works, and in fact you can permute the $m$ kids in the chain and it will still work.  There is enough cancellation that you will always end up with (*).

*Yes IMHO the problem should be approached by directly using (*), which is a generalization of the Binomial method.  Your method is just re-deriving / proving it.  :)

*I don't understand your 3rd question.  But whatever it's asking, I hope (*) answered it!
