Why the distinguishability changes the probability? Place $n$ distinct particles in $N$ ordered buckets where $N≥n$. The total number of possible placements is $N^n$. Choose $n$ particular buckets, then the probability of each of the $n$ bucket containing exactly one particle is $\frac{n!}{N^n}$.
Now suppose the particles are indistinguishable, then the number of possible placements is $C_{n+N-1}^n$ (denotes $n+N-1$ chooses $n$). Given the chosen $n$ buckets, the probability of each of the $n$ buckets containing exactly one particle is $\frac{1}{C_{n+N-1}^n }$.
Suppose $N=10, n=5$, then the first probability is $5! / 10^5 = 0.0012$, and the second probability is $1/C_{14}^5 \approx 0.0005$.
This is very counter-intuitive. The only difference between the two cases is if the particles are distinguishable. Anyone can help explain why the distinguishability changes the probability?
 A: Let us deal with a simple case, three particles and three buckets.
Consider two observers who are watching, one on a colour monitor, and the other on a carefully-contrasted black-and-white monitor, the same balls fall randomly into buckets and recording the results.   The colour monitor can distinguish the red, blue, and green balls, while the black-and-white cannot, and yet they are the same balls.   What is the probability that the balls fall into different buckets?
Now the first observer can table a total of $3^3$ possible outcomes, of these $3!$ the particles in different buckets.
The other observer can only view $^5C_3$ distinct outcomes, of which only one distinct way places one particle in each bucket.
Why will the probability both measure be closer to $3!/3^3$ rather than $1/{^5C_3}$ ?   Well, whether or not the particles can be distinguished by an observer, they still individual entities.   So the applicable model is: "each individual particle has an unbiased choice of one of three individual locations". 
$$\begin{array}{c|ccc}*|*|* & r|b|g & r|g|b & b|r|g & b|g|r & g|r|b & g|b|r 
\\ **||* & rb||g & rg||b & bg||r \\ **|*| & rb|g| & rg|b| & bg|r| \\ |**|* & |rb|g & |rg|b & |bg|r \\ |*|** & |r|bg & |b|rg & |g|rb \\ *|**| & r|bg| & b|rg| & g|rb| \\ *||** & r||bg & b||rg & g||rb
\\ *** || & rbg||
\\ |***| & |rbg|
\\ ||*** & ||rbg 
\end{array}$$

tl;dr The distinct arrangements of indistinct particles are not each equally probable states.
A: when you consider the particles as distinguishable you are specifying the state as an n-tuple whose $k^{th}$ element is the box number of the $k^{th}$ ball.
There are $N^n$ of these micro-states and they are all equally probable. 
When you think of the particles as indistinguishable you are specifying the the state as an N-tuple whose elements sum to n.  The $p^{th}$ element is the number of of balls in the $p^{th}$ box.   You are then creating macro-states, whose probability will be proportional to the number of micro-states corresponding to that macro-state.
e.g. 
case 1: there are $n!$ micro-states corresponding to the macro-state in which the first $n$ boxes contain 1 ball 
case 2: there is only  1 micro-state corresponding to the macro-state in which all $n$ balls are contained in the first box.  
both cases are described by a single macro-state 
but case 1 is more probable by a factor of $n!$
if $N=5$ and $n=2$ 
 the macro-state for case 1 is $(1,1,0,0,0)$ corresponding to the micro-states $(1,2)$ and $(2,1)$
 the macro-state for case 2 is $(2,0,0,0,0)$ corresponding to the micro-state $(1,1)$
