Pens Distributing problem If I were to distribute 31 DIFFERENT pens among 5 students, and each of them must at least receive 4 pens. How many different arrangements are there?
As I think it involves lots of different cases when the pens are different. I start from finding how many ways are there if the pens are identical.
Let A1 be the number of pens the first student get and A2 be the number of pens the second student get.... We have:
A1 + A2 + A3 + A4 + A5 = 31.
Let B1 = A1 - 4
B1 + B2 + B3 +B4 + B5 = 11
Which is equal to picking 4 spot out of 15 to put the boundaries separating the students and the pens. So I have 15C4 as the answer to IDENTICAL pens. I am stuck after that and not quite sure how to proceed. Can anyone help me out? I really appreciate any help you can provide.

I'll give you an hint. 
Each student compulsorily has 4 pens each. This leaves you with 11 extra pens to be distributed among the 5 people.
There are C(11,31) possibilities for these pens, which can be distributed in the following ways-


*

*only one of them gets all the  11 pens (this can be done in 5 ways)

*the 11 extra pens are distributed to 2 of them. 


(There are 2.C(2,5).[C(1,11)+C(2,11)+C(3,11)+C(4,11)+C(5,11)]= 43560 ways under this)


*

*the pens are distributed to 3 of them

*the pens are distributed to 4 of them

*the pens are distributed to 5 of them
Evaluate the number of ways in each case, add these values and multiply by C(4,20).C(4,16).C(4,12).C(4,8) . (the remaining 20 pens can be distributed to the 5 people such that each of them gets 4 in these many ways.)
