For counting permutations with identical objects, why does dividing nPr by the factorial of the number of identical objects give the correct answer? I can find plenty of sites that say that this works, but I can't seem to find an explanation for why it works. I'm rather stumped.
 A: Suppose you have n positions need to be filled by n items divided into m categories. Items in the same category are non-distinguishable. Let $x_1,x_2,...x_m$ denote number of items in each category, so you have $n=x_1+x_2+...+x_m$.
There are $n\choose x_1$ ways to choose positions for items in category 1.
There are ${n-x_1\choose x_2}$ ways to choose positions for items in category 2.
There are ${n-x_1-x_2\choose x_3}$ ways to choose positions for items in category 3.
...
Finally there are ${n-x_1-x_2...-x_{m-1}\choose x_m}$ ways to choose positions for items in category m.
Number of ways of permutation should be: ${n\choose x_1}\times {n-x_1\choose x_2}\times {n-x_1-x_2\choose x_3} \times \cdots \times{n-x_1-x_2...-x_{m-1}\choose x_m}=\frac{n!}{x_1!\times(n-x_1)!}\times \frac{(n-x_1)!}{x_2!\times(n-x_1-x_2)!}\times \frac{(n-x_1-x_2)!}{x_3!\times(n-x_1-x_2-x_3)!}\times\frac{(n-x_1-x_2..-x_{m-1})!}{x_m!\times0!}=\frac{n!}{x_1!\times x_2!\times...x_m!}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ 
A: Logical Answer:
Assume, n is the number of identical objects.
Dividing by n! is like saying that we are dividing by all the possible arrangements that might have occurred but they did not as the objects are identical. So, dividing by n! essentially rules out all such possibilities that did not exist.
