We usually distribute smartboards instead of blackboards these days!
We have a stars and bars problem.
We represent the blackboards by n=8 stars, and divide into k=4 schools (A,B, C & D) using three bars. Two consecutive bars or a bar at the end means that school has no blackboards.
For example,
*|***|*|***
Means there are two schools with 1 blackboard, and two other ones with three blackboards. Another example:
|*||*******
will mean there are two schools (A &C) with no boards, and B will have one, while school D will have 7.
So the problem reduces to counting the number of ways we can put the three identical bars.
If there is no minimum number of boards for each school, then we can place the k=4 bars at n+1=8+1=9 locations each.
Using combination notation, there are
C(n+1, k-1)=C(8+1,4-1)=C(9,3)=9!/(3!(9-3)!)=84 ways of distributing 8 blackboards to 4 schools.
On the other hand, if there is a minimum of one board for each school, then each school will get one, leaving 4 to be distributed as above, in
C(4+1,4-1)=C(5,3)=5!/(3!(5-3)!)=10 ways.
