The formula is (n+r-1)C(r-1) with n=10-3*2 = 4 whiteboards left after giving
each school 2, r = 3, the number of schools.  (4+3-1)C(3-1) = 6C2 = 15.

Here's why that works:

Let the schools br A, B, and C.

First give each of the 3 schools 2 whiteboards each, which guarantees
that each school will get at least 2 whiteboards.

Then there are 4 whiteboards left to distribute to one, two or all 3 
of the schools in addition to the 2 they already have. 

To figure this out, get (or imagine) 4+3= 7 slips of paper.  Lay them
left to right.  Write "School C" on the slip of paper on the far
right.

Now choose any 2 of the other 6 sheets of paper.  Write School A on the
slip of paper you just chose which is farther to the left, and write 
"School B" on the other one you chose, which is farther to the right.
You can do this in 6C2 = 15 ways.  This is the answer to the problem.
Why: OK, here is why:

You have 7 slips of paper laid out left to right.  The rightmost one has 
"School C" written on it.  One of the other slips of paper have "School A"
written on it. And one between it and the right end has "School B" written on
it.  The remaining 4 slips of paper have nothing written on them.  The number
of slips of paper to the left of the slip that reads "School A"  is the
number of whiteboards to give to school A in addition to the 2 it already
has.  (If the slip that reads School A" is on the left end, then there are no
slips of paper left of it.  Then the number of slips of paper left of it is 0,
so in that case School A gets 0 or NO additional whiteboards.

The number of slips of paper between the slips that reads "School A" and
"School B" is the number of whiteboards to give to school B in addition to the 2
it already has.  (If the slip that reads School B" happens to be adjacent to the
slip that reads "School A", then there are no slips of paper between them.  Then
the number of slips of paper between them is 0, so in that case School B gets 0
or NO additional whiteboards.
  
The number of slips of paper between the slips that reads "School B" and
"School C" on the right end is the number of whiteboards to give to school C in
addition to the 2 it already has.  (If the slip that reads School B" happens to
be adjacent to the slip that reads "School C", then there are no slips of paper
between them.  Then the number of slips of paper between them is 0, so in that
case School C gets 0 or NO additional whiteboards.

Each choice of those 2 slips of paper determines a different partition of of
the remaining 4 whiteboards.

Answer:  (n+r-1)C(r-1), n=10-3*2 = 4 (4+3-1)C(3-1) = 6C2 = 15.

Edwin