Probability - books on shelf

[jamesmathy]

we have 15 books - 9 (different) novels and 6 (different) math books.
what is the probability that the novels would be placed on a shelf in 2 groups of 6 and 3 with 1+ math books in between.

I know that there are 15! possible ways to place them all but im kinda lost with how many different group there are....

any help is welcome! thanks!

 

[Deleted member 4993]

we have 15 books - 9 (different) novels and 6 (different) math books.
what is the probability that the novels would be placed on a shelf in 2 groups of 6 and 3 with 1+ math books in between.

I know that there are 15! possible ways to place them all but im kinda lost with how many different group there are....

any help is welcome! thanks!

This - I think - is a nasty complicated problem with too many "choosings" and groups. Above my pay-grade.....

 

[Steven G]

As always you need to understand the event. You will have have 6 novels bundled together (think of them as 1 book) and the remaining 3 novels bundled together. In how many ways can you make these two bundles? There are 9! ways of arranging the 9 novels and all you need to do is partition them either between the 3rd and 4th book OR between the 6th and 7th book. This can be done in 2 ways.

Now where can the 6 math books go? Some can go to the left of the novels, some can go to the right of the novels and at least one must go in the middle of the novel. So the questions is how can you partition the 6 math books into three regions where the middle region must get (at least)1. This part reduces to how many ways can we place 5 math books into 3 different places given that you chose one to go in between the novels (in how many ways can that be done). This part you try on your own.

 

[pka]

we have 15 books - 9 (different) novels and 6 (different) math books.
what is the probability that the novels would be placed on a shelf in 2 groups of 6 and 3 with 1+ math books in between.

I find this to be a poorly worded question. It seems to me that the mathematics texts are problematic.
I assume the phrase "1+ math" means at least one or more".
The integer six can be partitioned into three nonempty cells: $411,~321,~\&~222$.
Those partitions can be in three, six, or one way.
Can you carry on?

 

[Dr.Peterson]

we have 15 books - 9 (different) novels and 6 (different) math books.
what is the probability that the novels would be placed on a shelf in 2 groups of 6 and 3 with 1+ math books in between.

I know that there are 15! possible ways to place them all but im kinda lost with how many different group there are....

I don't think you copied the exact wording of the problem, which is something we beg you to do, to avoid misunderstandings:

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My understanding is that it is intended to be something like this:

We have 9 novels and 6 math books, each distinct, that we want to place in one row on a shelf. If we place them randomly (as if we had shuffled them), what is the probability that 6 of the novels will be together as a group, and the other 3 in another group, with at least one math book separating them (and others possibly at either end)?​

That would be a whole lot clearer!

Along Jomo's lines, we can first arrange the 9 novels in a row (how many ways?), then split them into either 6+3 or 3+6 (2 choices), then arrange the 6 math books in a row (how many ways?), and then (as pka said, but with more possibilities as I understand it) split those into three parts, where the middle part must have at least 1, but the two outer parts could be empty. This might be handled by a stars and bars model.

I think we would get the same answer if we considered the books of each type indistinguishable, just modeling them like mmNNNNNNmmmNNNm.

 

[pka]

I don't think you copied the exact wording of the problem, which is something we beg you to do, to avoid misunderstandings:

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to help...

www.freemathhelp.com

Amen, Amen, & again AMEN

(as pka said, but with more possibilities as I understand it) split those into three parts, where the middle part must have at least 1, but the two outer parts could be empty. This might be handled by a stars and bars model.

I now think I was mistaken inn think that each of the three special areas needed to be non-empty. I now think that only the 'middle' need be not empty.

split those into three parts, where the middle part must have at least 1, but the two outer parts could be empty. This might be handled by a stars and bars model. I think we would get the same answer if we considered the books of each type indistinguishable, just modeling them like mmNNNNNNmmmNNNm.

I agree that using a generic string $NNNNNNNNNmmmmmm$ to model this makes so much more sense.
That string can be arranged in $\dfrac{15!}{9!\cdot 6!}=\dbinom{15}{6}=\text{fifteen choose six}=5005$.
In the spirit of dearly departed Soroban, duct-tape six of the $N's$ and three of the $N's$ together.
Now we have the string $\boxed{NNNNNN}\;\boxed{NNN}mmmmmm$ to consider.
That revised string can be arranged in $\dfrac{8!}{6!}=\dbinom{8}{6}=\text{eight choose six}=28$.
Six separators create seven places. We do not want the string $\boxed{NNNNNN}\;\boxed{NNN}$ in any of those places.
We need at least one $m$ between those two. So we can have $21$ places
That gives $\dfrac{21}{5005}=0.004195804195804195804195804195804195804195804195804195$

 

[Cubist]

Six separators create seven places. We do not want the string $\boxed{NNNNNN}\;\boxed{NNN}$ in any of those places.
We need at least one $m$ between those two. So we can have $21$ places
That gives $\dfrac{21}{5005}=0.004195804195804195804195804195804195804195804195804195$

I think $\boxed{NNNNNN}\;\boxed{NNN}$ could also be $\boxed{NNN}\;\boxed{NNNNNN}$ so the actual probability is double the above, i.e it ought to be 42/5005

I'm not sure we should be providing full answers like this.

 

[jamesmathy]

We have 9 novels and 6 math books, each distinct, that we want to place in one row on a shelf. If we place them randomly (as if we had shuffled them), what is the probability that 6 of the novels will be together as a group, and the other 3 in another group, with at least one math book separating them (and others possibly at either end)?

this is the right phrasing I believe (was not written originally in English)

 

[JeffM]

I think $\boxed{NNNNNN}\;\boxed{NNN}$ could also be $\boxed{NNN}\;\boxed{NNNNNN}$ so the actual probability is double the above, i.e it ought to be 42/5005

I'm not sure we should be providing full answers like this.

When the original problem is so badly stated, it is highly probable that the effort to determine the true problem ends up giving the solution. I am not saying that is right, but when people give incorrect answers or confusing answers due to misunderstanding the problem, their corrections and clarifications tend to give away more and more of the answer.