Combination with repetition

**mybrohshi5** · Oct 17th 2011, 03:28 PM

8 identical blackboards are divided among 4 schools:

1) what are the # of divisions possible?

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

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

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

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

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:

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

Thanks for any input

**Plato** · Oct 17th 2011, 03:40 PM

Originally Posted by mybrohshi5

8 identical blackboards are divided among 4 schools: 1) what are the # of divisions possible?
using combination with repetition formula $\binom{n+k-1}{n-1}$ I get $\binom{8+4-1}{4-1} = \binom{11}{3}$

2) how many divisions are possible if each school receives at least 1 blackboard?
$\color{red}\binom{8-1}{4-1} = \binom{7}{3}$

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

**mybrohshi5** · Oct 17th 2011, 03:44 PM

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...

**Plato** · Oct 17th 2011, 03:53 PM

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.

**mybrohshi5** · Oct 17th 2011, 03:59 PM

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.

**Plato** · Oct 17th 2011, 04:18 PM

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.

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