Combination with repetition

8 identical blackboards are divided among 4 schools:

1) what are the # of divisions possible?

using combination with repetition formula $\displaystyle \binom{n+k-1}{n-1}$

I get $\displaystyle \binom{8+4-1}{4-1} = \binom{11}{3}$

2) how many divisions are possible if each school receives at least 1 blackboard?

$\displaystyle \binom{8-1}{4-1} = \binom{7}{3}$

NOW FOR MY QUESTION:

I was curious to what would happen if it was asked: how many divisions are possible if each school receives at least 2 blackboards?

would it just be:

$\displaystyle \binom{8-2}{4-1} = \binom{6}{3}$

Thanks for any input :D

Re: Combination with repetition

Quote:

Originally Posted by

**mybrohshi5** 8 identical blackboards are divided among 4 schools: 1) what are the # of divisions possible?

using combination with repetition formula $\displaystyle \binom{n+k-1}{n-1}$ I get $\displaystyle \binom{8+4-1}{4-1} = \binom{11}{3}$

2) how many divisions are possible if each school receives at least 1 blackboard?

$\displaystyle \color{red}\binom{8-1}{4-1} = \binom{7}{3}$

The answer for 2) should be $\displaystyle \binom{4}{3}$

Re: Combination with repetition

Not to question your knowledge, but are you sure?

My probability solutions guide (and the back of my book) says it should be 35 which is 7 choose 3...

Re: Combination with repetition

Quote:

Originally Posted by

**mybrohshi5** Not to question your knowledge, but are you sure? My probability solutions guide (and the back of my book) says it should be 35 which is 7 choose 3...

Look, if each school gets at least one then that leaves only four to give out. Whoever wrote the solutions manual is just simply wrong.

Re: Combination with repetition

That sounds good to me. Thank you!

So if it was at least 2 blackboards given to each school, that means that NO blackboards are left to give out so would that just be 1?

And do you mind if I ask you about a similar problem dealing with people exiting an elevator?

I am studying for my midterm exam tomorrow and am really trying to understand combination problems.

Re: Combination with repetition

Quote:

Originally Posted by

**mybrohshi5** And do you mind if I ask you about a similar problem dealing with people exiting an elevator?

Not at all.

But by forum rules post in a new thread.

Re: Combination with repetition

Quote:

Originally Posted by

**Plato** Look, if each school gets at least one then that leaves only four to give out. Whoever wrote the solutions manual is just simply wrong.

This is an old post, but since it is the second option that came up when I googled the question, I think this merits a correction.

The question says that each school receives at least 1 blackboard. The other four are distributed in any way, so 1 school could end up with 5 and the others with just one, or each school could end up with 2 each, or any other distribution. That is why the solution is 7 choose 3, or 35.

4 0 0 0

3 1 0 0

3 0 1 0

3 0 0 1

etc...