By: Rich Zwelling, Apex GMAT Instructor
GMAT Combinatorics. It’s a phrase that’s stricken fear in the hearts of many of my students. And it makes sense, because so few of us are taught anything about it growing up. But the good news is that, despite the scary title, what you need to know for GMAT combinatorics problems is actually not terribly complex.
To start, let’s look at one of the most commonly asked questions related to GMAT combinatorics, namely the difference between combinatorics and permutations:
Does Order Matter?
It’s important to understand conceptually what makes permutations and combinations differ from one another. Quite simply, it’s whether we care about the order of the elements involved. Let’s look at these concrete examples to make things a little clearer:
Suppose we have five paintings to hang on a wall, and we want to know in how many different ways we can arrange the paintings. It’s the word “arrange” that often gives away that we care about the order in which the paintings appear. Let’s call the paintings A, B, C, D, and E:
Each of the above three is considered distinct in this problem, because the order, and thus the arrangement, changes. This is what defines this situation as a PERMUTATION problem.
Mathematically, how would we answer this question? Well, quite simply, we would consider the number of options we have for each “slot” on the wall. We have five options at the start for the first slot:
_5_ ___ ___ ___ ___
After that painting is in place, there are four remaining that are available for the next slot:
_5_ _4_ ___ ___ ___
From there, the pattern continues until all slots are filled:
_5_ _4_ _3_ _2_ _1_
The final step is to simply multiply these numbers to get 5*4*3*2*1 = 120 arrangements of the five paintings. The quantity 5*4*3*2*1 is also often represented by the exclamation point notation 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)
So, what about COMBINATIONS? Obviously if we care about order for permutations, that implies we do NOT care about order for combinations. But what does such a situation look like?
Suppose there’s a local food competition, and I’m told that a group of judges will taste 50 dishes at the competition. A first, a second, and a third prize will be given to the top three dishes, which will then have the honor of competing at the state competition in a few months. I want to know how many possible groups of three dishes out of the original 50 could potentially be selected by the judges to move on to the state competition.
The math here is a little more complicated without a combinatorics formula, but we’re just going to focus on the conceptual element for the moment. How do we know this is a COMBINATION situation instead of a permutation question?
It’s a little tricky, because at first glance, you might consider the first, second, and third prizes and believe that order matters. Suppose that Dish A wins first prize, Dish B wins second prize, and Dish C wins third prize. Call that ABC. Isn’t that a distinct situation from BAC? Or CAB?
Well, that’s where you have to pay very close attention to exactly what the question asks. If we were asking about distinct arrangements of prize winnings, then yes, this would be a permutation question, and we would have to consider ABC apart from BAC apart from CAB, etc.
However, what does the question ask about specifically? It asks about which dishes advance to the state competition? Also notice that the question specifically uses the word “group,” which is often a huge signal for combinations questions. This implies that the total is more important than the individual parts. If we take ABC and switch it to BAC or BCA or ACB, do we end up with a different group of three dishes that advances to the state competition? No. It’s the same COMBINATION of dishes.
It’s interesting to note that there will always be fewer combinations than permutations, given a common set of elements. Why? Let’s use the above simple scenario of three elements as an illustration and write out all the possible permutations of ABC. It’s straightforward enough to brute-force this by including two each starting with A, two each starting with B, etc:
But you could also see that there are 3*2*1 = 3! = 6 permutations by using the same method we used for the painting example above. Now, how many combinations does this constitute? Notice they all consist of the same group of three letters, and thus this is actually just one combination. We had to divide the original 6 permutations by 3! to get the correct number of permutations.
Next time, we’ll continue our discussion of permutation math and begin a discussion of the mechanics of combination math.
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