By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 2021
Review of example from last post
Last time, when we started our discussion of GMAT Combinatorics, we gave a brief example of GMAT permutations in which we had five paintings and asked how many arrangements could be made on a wall with those paintings. As it turns out, no complicated combinatorics formula is necessary. You can create an easy graph with dashes and list five options for the first slot, leaving four for the second slot, and so on:
_5_ _4_ _3_ _2_ _1_
Then multiply 5*4*3*2*1 to get 120 arrangements of the five paintings. Remember you could see this notationally as 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)
More permutation math
But there could be fewer slots then items. Take the following combinatorics practice problem:
At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge?
First, as a review, how do we know this is a PERMUTATION and not a COMBINATION? Because order matters. In the previous problem, the word “arrangements” gave away that we care about the order in which items appear. In this problem, we’re told that we’re interested in the “orderings” of cheeses. Cheddar followed by gouda would be considered distinct from gouda followed by cheddar. (Look for signal words like “arrangements” or “orderings” to indicate a PERMUTATION problem.)
In this case, we must consider the options of three or four cheeses separately, as they are independent (i.e. they cannot both happen). But for each case, the process is actually no different from what we discussed last time. We can simply consider each case separately and create dashes (slots) for each option. In the first case (three cheeses), there are six options for the first slot, five for the second, and four for the third:
_6_ _5_ _4_
We multiply those together to give us 6*5*4 = 120 possible ways to present three cheeses. We do likewise for the four-cheese case:
_6_ _5_ _4_ _3_
We multiply those together to give us 6*5*4*3 = 360 possible ways to present four cheeses.
Since these two situations (three cheeses and four cheeses) are independent, we simply add them up to get a final answer of 120+360 = 480 possible orderings of cheeses, and the correct answer is D.
You might have also noticed that there’s a sneaky arithmetic shortcut. You’ll notice that you have to add 6*5*4 + 6*5*4*3. Instead of multiplying each case separately, you can factor out 6*5*4 from the sum, as follows:
6*5*4 + 6*5*4*3
= 6*5*4 ( 1 + 3)
= 30*16 OR 20*24
Develop the habit of looking for quick, efficient ways of doing basic arithmetic to bank time. It will pay off when you have to do more difficult questions in the latter part of the test.
Now that we have been through GMAT permutations, next time, I’ll give this problem a little twist and show you how to make it a COMBINATION problem. Until then…
Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches