Below post is from my comment

@ http://www.cat4mba.com/node/6201

updated here as I thought it might help others

Let’s make the question a bit simple. Say we have 6 similar balls and we want to distribute it among 3 persons where each person should get at least one ball.

Imagine that the 6 balls are placed somewhere and we want to divide it into three groups by putting two separators (marks) in between them

This division can be made in many different ways like
 

Now think what exactly we are doing – we are putting 2 different marks in 5 different places. How many combinations are possible?

Isn’t it ⁵C₂?

Now suppose we did not have the restriction of giving at least one ball to each person.

Then we can put the marks in either end and so total number of places becomes 7 (6 + 2 -1)
So its permutation of 7 different items where 2items are of type 1 and 5 are of type 2.

Possible number of arrangements = 7! / 5! 2!

In general,

If n identical objects are to be distributed among m persons then the number of possible ways of doing it is

^n-1C_m-1  when each person should get at least one object

^n+m-1C_m-1 w/o any restriction.