look at every student separately

first student we can choose 4 object from 20 so 20C4

second student we only have 16 objects left so 16C4

and so on

final answer being 20C4*16C4*12C4*8C4*1

The line of thought for B is as follows:

if any number can be distributed => all of numbers can be chosen from the set. not you are not dealing with only one distribution but many hence sum applies in a manner of :

20C0+20C1+20C2+20C3+20C4.......


of course you know we have symmetry here (20C1=20C19 and 20C2=20C18 and so on)

this will save you some crunching

For (a), the answer given by dimonya looks correct. But for (b) am not sure.

There are some formula you can remember.

1) No of way in which 'n' distinct objects can be distributed over 'k' bins in any order = k^n

2) The no of ways in which 'n' distinct object can be distirbuted over 'k' bins such that all bins contain equal number of objects is n! / [(n/k)!^k]

3)The no of ways in which 'n' "identical" object can be distributed over 'k' bins in any order is = C[(n+k-1),(k-1)]

The difference between 1 and 3 are that, in 1 objects are not identical.

Now for question (a), we use the 2nd formula

Here n=20, k = 5 and n/k = 20/5 = 4

So we have the no of ways as = (20!)/[(4!)^5].

For (b) we need to use the 1st formula.

Here n=20 and k=5, so we have 5^20.

Its difficult to explain the way these formula are derived. You may want to check this document if you really want to know (https://books.google.com/books?id=qCBPSN ... t#PPA45,M1) Page 45 - 55