Ordering Books On a Shelf 
The are 6 different finance books and 4 different math books, there were arranged randomly, what is the probability that books from the same subject are standing one by the other.

The answer is fairly simple there are $10!$ ways to order 10 books, $6!$ to order the finance one $4!$ and 2 more for swapping sides (left<>right).
So the answer is ${6!*4!*2\over 10!}=\frac {2}{10 \choose 4}$, which is equivalent to the answer of the same question when there are 6 finance books that are the same  and 4 math books that are the same. Why is that?
 A: *

*Number of finance books=6

*number of maths books=4
 - 



step1

consider all the 

different  finance books 

as 1unit 


*

*and 


all the 

different maths books

as 1 unit 
According to your question 
Now 
we are having just two different units ( first one is finance books unit and the second one is maths books unit )


*

*these 2 different units can be arranged  in them selves in 2! Ways 


And 
 that finance books are 

each is "different to the other "
  We can arrange them , them selves in 6! Ways.... Here we are having different arrangements with books  (
  if even in case "that finance books are same books" then the resulting arrangements should be 6!/6!  =1 way)

Next 


*

*similarly the different maths books are arranged them selves in 4! Ways
(If even in case they are same maths books then the arrangements are 4!/4!=1 way)


Now probability=2!(6!)(4!)/10! 
Is the answer 
If books are same I mean finance books are not different they are same and maths books are same 
Then prob={[2!(1)(1)]/[10!/4!6!]}
A: Suppose the finance books are identical and so are the finance books. No matter how we order them, there are always $6!\cdot4!$ ways to reorder them so that the math books occupy the same places they did before reordering them.
In other words, each order when the books are identical gives us exactly $6!\cdot 4!$ orders when the books are distinct, so this does not depend in any way in which order we choose, it is the same for every one. It is also true that the only orders that work are the $2\cdot6!\cdot4!$ that result from the two unordered orders that work.
In conclusion: the reason it works is that all the old (all books identical) unordered arrangements give place to an equal number of new arrangements (all books distinct) , so it does not depend on the specific old arrangement.
