Problem Solving

A conference is to have eight presentations over the course of one day, consisting of three long presentations, and five short presentations. If the conference organizer doesn't want consecutive long presentations, and the conference is to start with a short presentation, how many schedules of presentations are possible?
120
720
2880
5760
11520

We can use the slot method here, but it needs to be looked at in a certain way.

First I want to consider ONLY the order of the short presentations (pretend there are no long presentations).

Well, there are 5 short presentations, so we can arrange them in 5! ways.

Now let's add the 3 long presentations.

To ensure that 2 long presentations are never consecutive, consider the following options (spaces) for long presentations:

short ___ short ___ short ___ short ___ short ___

The 5 short presentations (shown) can be arranged in 5! ways.
We can now place the 3 long presentations in any of the 5 spaces shown. This will ensure that 2 long presentations are never consecutive.

Let's let A, B and C be the 3 long presentations to add to the schedule.

Schedule presentation A: there are 5 slots, so we can schedule presentation A in 5 ways.
Schedule presentation B: there are 4 slots remaining, so we can schedule presentation B in 4 ways.
Schedule presentation C: there are 3 slots remaining, so we can schedule presentation c in 3 ways.

The total number of ways to schedule all 8 presentations = 5! x 5 x 4 x 3 = 7200
So, the answer is not listed

Cheers,
Brent

jsnipes wrote: The book's explanation says:
Let S represent a short presentation, and L represent a long presentation. The schedule must start with an S, and every L must be followed by at least one short presentation. So the schedule must be SLSLSLS, along with one more short presentation inserted somewhere. There are four ways to do this; SSLSLSLS, SLSSLSLS, SLSLSSLS, and SLSLSLSS. For each of these there are 5! arrangements of the short presentations, and 3! arrangements of the long presentations, giving 120x6=720 possible schedules for each ordering. So there are 4x720=2,880 total possible schedules.

Hmmm, the solution in the book excludes arrangements that end with a long presentation.
We could also have:
SSSLSLSL
SLSSSLSL
SLSLSSSL

Since there's nothing that prohibits us from ending the day with a long presentation, the author(s) of the book missed some possibilities.

Cheers,
Brent

The book also missed:
SLSSLSSL
SSLSLSSL
SSLSSLSL

Cheers,
Brent

jsnipes wrote:Yea, it definitely did miss all of those. Frustrating when you buy a book and it has bad/vague questions or wrong answers. Appreciate the help on this one. Do you think that the slot method is still the best way to approach this problem if you saw it on a test? Or how would you best attack this problem?

I always begin by determining whether or not I can use the slot method (aka: FPC). I would have used it for this question.

Cheers,
Brent

jsnipes wrote:Brent, are there some obvious examples of when you CANNOT use the slot method?

The fast answer is "combinations." If a question involves combinations, then there typically isn't a nice way to use the slot method.

I recently wrote two articles about the slot method (Fundamental Counting Principle) at:
- https://magoosh.com/gre/2011/gre-combina ... binations/
- https://magoosh.com/gre/2011/gre-combina ... s-part-ii/

That may help you determine when you can use the slot/FCP method.

Cheers,
Brent