How many ways can you distribute blackboard to schools?

Question

We have 8 indistinguishable blackboards that we wish to distribute over 4 schools in how many ways can we distribute them?

is it the same as saying how many ways can we put 8 objects in 32 slots (every school is 8 slots)?

but when i saw the solution to the problem the answer was: $\frac{11!}{3!*8!}$ and i couldn't figure what those numbers signify

This method is often known as "stars and bars," as it is counting the number of ways to place $n$ bars in between $m$ (indistinguishable) stars: $$\ast \mid \ast \ast \mid \ast\ast\ast\ast$$ — While I Am, Jan 16 '22 at 23:30
See this "abstract duplicate" for the stars-and-bars method in combinatorics. — hardmath, Jan 20 '22 at 00:52

score 2 · Accepted Answer · answered Jan 16 '22 at 22:44

If I understand your question correctly: Each (distinct)school could have different number of indistinct blackboard, and the total number of blackboards is 8. For instance, school A,B,C,D could have 1,2,4,1 blackboards respectively.

You could essentially set x,y,z,w to be the number of blackboards owned by the 4 schools. Then we have integers x,y,z,w satisfying: $$0\leq x,y,z,w\leq 8$$ $$x+y+z+w = 8$$

The number of distribution is then number of possible solutions of (x,y,z,w). (Assuming you are familiar with ${n\choose k} = \frac{n!}{k!(n-k)!} $ for $0\leq k \leq n$)

The number of possible solutions of (x,y,z,w) can be calculated by $${(8+3)\choose 3 }$$ The idea behind this is that you consider number of permutation of eight B's and 3 +'s, each B represent a black board, and + a plus sign.

For instance a solution (x,y,z,w)=(1,2,4,1) would correspond to an arrangement of B's and +'s, specidically B+BB+BBBB+B

and solution (x,y,z,w)=(7,1,0,0) correspond to BBBBBBB+B++

Then it is clear that the each of arrangement of 8 B's and 3 +'s correspond to exactly one solution of $x+y+z+w = 8$, with each x,y,z,w being non-negative. So the number of solutions (x,y,z,w ) = number of arrangement of 8 B's and 3 +'s, which is $${8+3\choose 3}= \frac{11!}{3!8!} (or {8+3\choose 8})$$

Out of total of 11 space, choose 3 out of 11 to be +, the rest has only the possibility of being B.

Note: the three comes from 4-1, out of x+y+z+w, there are 4-1 "+'s" in bewteen. Note2: the fact that x,y,z,w are non-negative is important. Since the arrangements can have + right next to another +, like the example of (x,y,z,w)=(7,1,0,0).

If you have restrictions like each school need to have at least 1 blackboard, then you can do the same thing by considering x'=(x+1), y'=(y+1)...w'=(w+1), where x' is the number of blackboard received by school A. Then list the same constrain, and use x, y, w, z that are non-negative, instead of x',y',z',w' which are now strictly positive.

No it is not the same as 8 things in 32 holes. For instance, if school A has 8 distance holes(A1, A2..., A8). Then the case of "A school having 2 blackboards" would have the possibility of it being in (A1,A2) or (A1,A3).... till (A7,A8) (total of ${8\choose 2} numbers in this case) So it is not the same

thank you for your clear answer, does this method of solving have a name it's the first time i see it ? — Rafik Bouloudene, Jan 16 '22 at 22:54
if i remember correctly it's called combination with repetition — fourofour_un, Jan 16 '22 at 23:25

How many ways can you distribute blackboard to schools?

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