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The Hidden Complexity of a Sandwich Order:
Fundamentals of Combinations and Permutations on the GMAT
Jeff Rogers, GMAT Specialist, Test Prep New York – San Francisco
Imagine yourself ordering a sandwich. For most of us, and most GMAT test takers, this isn’t much of a stretch. You’ve been there before. You’ve sidled up to the counter and made a set of snap decisions regarding the bread, meat, and cheese. You’ve selected your favorite sandwich option among any number of possible sandwiches. What you probably haven’t done is to stop and calculate just what that total number of possible sandwiches actually is.
To do that, you’d use what’s often called the fundamental counting principle, i.e., the idea that if you have a ways of doing one thing and b ways of doing something else, you have a x b ways of doing the two things together. So if you have 4 choices of bread, 5 choices of meat, and 6 choices of cheese at the sandwich counter, you’re actually choosing between 120 (or 4 x 5 x 6) possible sandwiches.
That sounds like a lot of options, but you’re still not looking at much of a real-world sandwich with just bread, meat, and cheese. Any sandwich shop worth its salt is probably going to offer some veggies. So let’s say you’re also looking at tomato, lettuce, onion, pickles, pepperoncinis, black olives, avocado, mushrooms, and spinach. That’s 9 possible veggies, and the shop offers any sandwich with your choice of 5 of the 9 veggies included.
Calculating the number of veggie combinations possible in choosing 5 items out of a 9-item pool is a little (but not much) more complicated than applying the fundamental counting principle. Anytime you’re choosing x items out of a y-item pool, you can apply the formula:
,ð‘¦!-ð‘¥!,ð‘¦âˆ’ð‘¥.!., which is simply taking the factorial of the total number of items and dividing it by the product of the factorials of the number of chosen items and the number of items not chosen. In our example of choosing 5 veggie options out of a 9-veggie pool, the formula would be, 9!-5!,9âˆ’5.!., which we can rewrite as 9 Ã— 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 -5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 ,4 Ã— 3 Ã— 2 Ã— 1... Cancelling a 5! in the numerator and denominator, we have, 9 Ã— 8 Ã— 7 Ã— 6 -, 4 Ã— 3 Ã— 2 Ã— 1.., which further reduces to 9 x 2 x 7 when we cancel out the remaining common factors from the denominator. Accordingly, we can choose 144 different combinations of 5 veggies from the 9-veggie list. And if we were considering the veggie decision in conjunction with the previously described sandwich decision, we’re now looking at 144 x 120 possible sandwiches. That’s 17,280 possibilities.
In our veggie combination example, the order of the veggies chosen for the sandwich obviously didn’t matter. Tomato, mushroom, avocado, black olive, and onion counts as the same topping set as mushroom, black olive, onion, avocado, and tomato. But there are other situations, of course, where the order of the items in the combination makes a difference. Imagine you’ve ordered 5 different sandwiches (A, B, C, D, and E) for your five co-workers. If each sandwich is going to one of your co-workers””who are lined up in a row, waiting for lunch””then it obviously makes a difference whether you distribute the sandwiches in the order A B C D E or C A D E B. In this case, where order matters, the total number of possible distributions of the five different sandwiches is given by the factorial ,ð‘¡ð‘œð‘¡ð‘Žð‘™ # ð‘œð‘“ ð‘–ð‘¡ð‘’ð‘šð‘ .! or, in this case, simply 5!. In other words, you can distribute the sandwiches 120 different ways. This is called a permutation, as distinct from the combinations discussed above.
Suppose, however, that your order came out wrong, and although you wound up with your 5 sandwiches, 2 of them were exactly the same. As you’d probably figure intuitively, that’s going to result in fewer distinct possible distributions among your 5 co-workers. In effect, you’ve wound up with A B C D D instead of A B C D E. And in the case of duplicates, the permutation is divided by the factorial of the number of duplicates, i.e., ,ð‘¡ð‘œð‘¡ð‘Žð‘™ # ð‘œð‘“ ð‘–ð‘¡ð‘’ð‘šð‘ !-ð‘‘ð‘¢ð‘ð‘™ð‘–ð‘ð‘Žð‘¡ð‘’ð‘ !.. Here, you’d have, 5!-2!., yielding a total of only 60 possible distinct sandwich distributions.
Or imagine that one of your 5 co-workers has decided to boycott the sandwich shop, and you return to the office with sandwiches A B C D E only to find out that he’s already eaten a salad brought from home. Now, you have a situation where you’ve got 5 different sandwiches that can be distributed among 4 people (i.e., you’re looking at the number of ways you can arrange 4 distinct items chosen from a 5-item set). Does order matter? Yes. But first you have to look at the number of different 4-sandwhich combinations you can pick from the 5-item set. So the formula this time is ,ð‘¡ð‘œð‘¡ð‘Žð‘™ # ð‘œð‘“ ð‘–ð‘¡ð‘’ð‘šð‘ !-ð‘â„Žð‘œð‘ ð‘’ð‘› ð‘–ð‘¡ð‘’ð‘šð‘ !(ð‘›ð‘œð‘¡ ð‘â„Žð‘œð‘ ð‘’ð‘›!). , or 5! divided by 4!, yielding only 5 possible 4-sandwhich combinations. Now, you can consider the number of distinct arrangements of each four-sandwich set. For each set, you have 4! arrangements. That’s 24 distributions for each of the five 4-sandwich combinations, or 120 possible ways of distributing four of five sandwiches to four different people.
Now, after slogging through all of this, you may never take ordering a sandwich quite as lightly again, but the upside is that you’ll be well versed in the fundamentals of combinations and permutations, two topics that tend to pop up frequently in higher level GMAT Quant questions. The GMAT loves combination and permutation questions because they’re a little counterintuitive for many test takers and because both subjects incorporate well into complicated word problems. That being said, you can tackle the majority of these questions with the simple formulas detailed above, provided that you read carefully, think critically, and set yourself up to answer the question the test is really asking.
So consider this a useful primer. Now get out to the sandwich counter and get yourself some practice.
Oh, and if you’re thinking of dessert, think of this. There’s an ice cream shop down the street from the sandwich place, and they’ve got a great 3-scoop waffle cone for just 25 cents. Why so cheap? Because the guy behind the counter picks the flavors at random from the 5 flavors of ice cream that the shop stocks.
The shop has Vanilla, Chocolate, Strawberry, Mint, and Pistachio. Every time I go in, I’m hoping for a combination of the first three. VCS, VSC, SCV, SVC, CSV, CVS. It’s vanilla, chocolate, and strawberry in any order (3!), and I like it. But I hate pistachio and mint. So when I think about going for a cone, I’m thinking about what the likelihood of getting the three flavors I like for three scoops chosen at random from a five-flavor set.
To figure that one out, I have to start by determining the number of 3-flavor combinations possible from a 5-flavor pool. That’s ,ð‘¡ð‘œð‘¡ð‘Žð‘™ # ð‘œð‘“ ð‘–ð‘¡ð‘’ð‘šð‘ !-ð‘â„Žð‘œð‘ ð‘’ð‘› ð‘–ð‘¡ð‘’ð‘šð‘ !. or in this case, ,5!-3!. , yielding 20 possible combinations. And I *only* like the combinations that are vanilla, chocolate, and strawberry””in any order. That’s 6, as we figured writing them out.
Life would be easier if I was less picky, but I’m not. So I have to look at the probability of getting a cone that I like when it’s going to be chosen from all of the possible cones out there. Using the general probability formula ,ð‘‘ð‘’ð‘ ð‘–ð‘Ÿð‘’ð‘‘ ð‘œð‘¢ð‘¡ð‘ð‘œð‘šð‘’ð‘ -ð‘¡ð‘œð‘¡ð‘Žð‘™ ð‘ð‘œð‘ ð‘ ð‘–ð‘ð‘™ð‘’ ð‘œð‘¢ð‘¡ð‘ð‘œð‘šð‘’ð‘ . , we can say that I’ll get a cone I’m happy with only 6 times out of 20. That’s 3 in 10 cones or a 30% chance that I’ll get the vanilla-chocolate-strawberry cone that I’m after.
Now the only remaining question: Am I willing to risk disappoint 70% of the time for 25 cents a pop? Absolutely.
© 2012, Test Prep New York
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