19
2012
“Never Fear, A Shortcut is Here!” Part II: Permutations
In my last blog entry, I wrote about combinations – it’s one of my favorite topics, and hopefully you’re feeling more confident about it after learning the tricks discussed. Today, let’s follow up with a look at how to tackle permutations, also known as “those terrible questions that ask me to arrange number and letters”.
Permutations may seem terrible, especially if you use the formula
, but they’re just as susceptible to a straightforward approach as are combinations. Let’s look at an example to demonstrate:
I set up permutations using the “slot method”: Line up as many “slots” as you are arranging items, and fill in each slot with the number of items that could possibly go in that place. In this case, we’re arranging 5 items, so I’ll set up 5 slots: ___ ___ ___ ___ ___
Since NEUTRAL has 7 letters, and any of them could go in the first space, we put a 7 there:
7 ___ ___ ___ ___
Once one letter is “placed” in that first slot, there are 6 remaining options for what can go in the second slot:
7, 6 ___ ___ ___
Hopefully you’re starting to see the pattern here – we’re just decreasing the number in each slot by 1, as there will be 1 fewer letter to “place” as each slot gets filled. Our completed chart looks like this:
7, 6, 5, 4, 3. To calculate the number of permutations, we just multiply each of these numbers together: 7*6*5*4*3 = 2,520, and answer choice (D) is correct.
The slot method applies no matter how many rules or restrictions a problem gives you on where certain items can be placed. Let’s look at a tougher example:
First, let’s ignore the restriction about David and Alice and see how many arrangements of 6 people there are: 6*5*4*3*2*1 = 720. Now we need to figure out how many of those arrangements include David and Alice standing next to each other. It’ll be easier here to figure out how many scenarios actually have them standing next to each other, and subtract that from the total of 720.
We can draw a couple of sketches to visualize:
Al, Da, ___ ___ ___ ___
___ Al, Da, ___ ___ ___
And so on, and so forth, until we see that there are a total of 10 scenarios in which David and Alice are next to each other: 5 in which Alice is to the left of David, and 5 in which David is to the left of Alice.
For each “placement” of David and Alice, we can’t forget to arrange the other 4 people:
Al, Da, 4* 3* 2* 1. In each instance in which these two are standing next to each other, there are 24 arrangements of the other people .
(10 scenarios in which David and Alice are next to each other) * (24 arrangements of the remaining 4 people) = 240 total “prohibited” permutations. (720 total arrangements) – (240 prohibited permutations) = 480 scenarios in which Alice and David are not next to each other, and Quantity B is greater.
Permutations provide a great opportunity to pick up points that the vast majority of test-takers leave on the table – use the slots to your advantage and see the impact that it has on your score!
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