The bookshelf approach is one of the coolest Permutations techniques going.

It can do the following

• Reduce what seems to be a crazy manual search for all possible combinations into a smooth formula


• Reinforce the notion of how "for every" works in Permutations


• Improve your librarian skills (OK, I'm kidding here)

But it's so cool, I have to show it to you anyway. Here goes!

Imagine you have a bookshelf, with ten different titles on it. How many ways can I rearrange the bookshelf so that it's a new arrangement each time? 10!! (or ten factorial exclamation mark )

Now, what if two of the ten titles are identical?... Hmmm... What's the impact there? Well clearly if I take any given permutation of the bookshelf, and swap the two identical titles around, ... it looks the same! So, if I want to count the arrangements of the bookshelf I have, then it's not as many as when they were all different.

What to do... what to do....

What I love about this problem is that when I ask my class what to do, they ALWAYS say the same thing, and they're ALWAYS wrong. But that's OK, cause I got it wrong too when I did this first.

They always say "Subtract the repetitions!". When I goad them into offering more detail on how many I should subtract, they say "Subtract 2!!" (by now you'll have cottoned onto my tongue in cheek usage of exclamation marks in permutation questions so further comment is superfluous. Such is the nature of tongue in cheek humour. Shut up, Micheal.)

This gets to the core of how "For every" works.

"For Every" snapshot of the bookshelf, there's an identical twin out there (in fact there are 2! copies of each snapshot) that looks the same. So there are 2! TIMES TOO MANY snapshots if we count them as being all different.

2! times too many? So we should .... DIVIDE BY 2! to reduce the repetition in the number, and stop overcounting! (that last exclamation mark was an exclamation mark. Not a factorial. Get on with it, Micheal. OK.).

So, the actual number of different ARRANGEMENTS of the bookshelf, where two identical titles are in the list are 10! / 2!.

Awesome.

Now what if there were 3 identical titles? Well, there would be 3! TIMES TOO MANY Permutations. So, divide by that.

Now ... WHAT IF ... I had two copies the OG and three copies of the GMAT Zone Roadmap in the 10 books? Well, you tell me.

There are 2! TIMES TOO MANY arrangements if we consider each OG as unique, and 3! TIMES TOO MANY arrangement if we consider each Roadmap as unique.

So the total number of unique arrangements are 10! / (2! 3!). I gotta reduce by the two repetitions.

Awesome. I've gained entry requirements to Librarianship 102 (given I've just passed muster for 101). I've proved a worthy foe for bookshelf overcounters. Call me Conan. The Librarian.

So where else can I use this? After all, I've got a rep to protect, as Travolta said in Grease. Can't be in the library all the time?

Good news : I can use it in the streets! Yes!

Where? How? Well, look for the problem in the Forum about getting from point A to point B using the shortest possible route.

Then while you at it, answer me this : how many arrangements of the letters in the word MEDITERRANEAN are possible when each new arrangement must start with E and end with N?

Again, answer in the Forum... coming soon... to a GMAT Zone class near you...