Combination with repetition

8 identical blackboards are divided among 4 schools:

1) what are the # of divisions possible?

using combination with repetition formula

I get

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

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:

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

I get

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

The answer for 2) should be

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.