Data Analysis
2.1 Arithmetic
 
2.2 Algebra
 
 
2.3 Geometry
 
2.4 Data Analysis
 
Data Analysis
If there’s one thing we can say about modern society, it’s that we love our data. If you spend just a few minutes on the Internet, you can find out roughly how many people there are in America; how much, on average, they earn; how long they’re expected to live; what percentage of them marry; what percentage have kids; how many kids they have; how much those kids weigh; and so on and so forth until every aspect of daily life is reduced to numerical form. Perhaps expressing everything numerically makes us feel in control of our fate, or perhaps it helps us cope with the ambiguities of our complex technological lives. Whatever the reason, we love our data, we surround ourselves with it, and we have invented all sorts of statistical mechanisms to express what all this information means.
The GRE test makers have taken our numerical fetish to heart. In this final Math 101 section, you’ll find more than you ever wanted to know about statistical concepts such as mean, median, mode, probability, and every other data analysis topic tested on the GRE.
Mean
On the GRE, mean and arithmetic mean both represent the concept that you may recognize by its more common name, average. No matter what we call it, the calculation is the same: Add up all the terms and divide by the number of terms. You’ve no doubt seen this in school: If you get scores of 90, 95, and 100 on three tests, then 95 is the average of the three test scores. In this basic example you can probably see at a glance that 95 is the average, but technically you can calculate it by taking the sum of the scores (90 + 95 + 100) and dividing it by the number of scores (3). The formula, in general terms, is:
Let’s try one out:
What’s the arithmetic mean of 3, –5, 7, and 0?
Solve by using the formula:
Some mean problems may be straightforward like the one above, but the more complicated ones may give you two values and ask you to solve for a third. For example, the test makers might give you the mean and the number of terms and ask you to solve for the sum of the terms. Your job will still be to plug the known values into the formula and solve from there. Here’s an example:

The average height of five people is 54 inches. One of the people leaves the group, and the average height of those remaining is 52 inches. How tall is the person who left?

In the first sentence, we’re given the number of people, five, and their average height, 54. We can use the mean formula to calculate the sum of the heights of these five people:
In the second sentence, we’re told that the average height of the remaining people is 52. Since one person left the group, four people remain. Plugging 4 and 52 into the mean formula gives:
The difference between the sum of the heights of the original five people and the sum of the heights of the remaining four must be equal to the height of the person who left. Subtract the second sum from the first to get the height of the person who left: 270 – 208 = 62 inches, the final answer.
Median
The median of a group of numbers is the middle term when the numbers are written in either ascending or descending order. That means that before you can calculate a median, you must first rewrite the terms of the group in ascending or descending order. For example, to calculate the median of 0.3, 7, 0, 9, and 10, you can’t choose 0 simply because it appears in the middle. You must first write the numbers in order: 0, 0.3, 7, 9, 10. Since 7 appears in the middle of this ordered list, 7 is the median.
If two numbers appear in the middle, which will happen whenever the total number of terms is even, take the mean of the two middle numbers to determine the median. For example, the median of 1, 2, 4, and 8 is 3, since 3 is the mean of 2 and 4.
One wrinkle you may come across in a median problem is a description of a list of consecutive numbers, instead of a list of the actual numbers, as in this example:

What is the median of all the integers between 210 and 260, inclusive?

You certainly don’t want to write out all the numbers from 210 to 260, and then try to find the one in the middle. It’s better to use the following formula:
In our example, the median would be:
Remember, inclusive means “including,” which is why we used 210 and 260. Had the question said exclusive, we would have used 211 and 259, divided by 2, and also gotten a median of 235.
Mode
The mode of a group of numbers is the number that occurs most frequently. If multiple numbers are tied for first place in the race to occur the most, then the group will have more than one mode. For example, the modes of the set of numbers {6, 6, 1, 3, 4, 1} are 6 and 1, since both 6 and 1 occur twice, and all the other numbers occur only once.
Range
The range of a group of numbers is the difference between the largest term and the smallest term. For example, the range of 10, –25, 3, 2, and 4 is 10 – (–25) = 35. One would need to travel 35 units on a number line to get from the smallest value, –25, to the largest value, 10.
Standard Deviation
Standard deviation is one of the most difficult statistical concepts, but thankfully you’ll only need a very general understanding of it for the GRE. The test makers won’t ask you to actually calculate standard deviation, as the formula for doing so is pretty difficult. You will, however, be expected to know that standard deviation is a measure of how spread out a group of numbers is. The more spread out a group of numbers, the larger its standard deviation. Let’s look at an example:

Which of the following groups of numbers has the greater standard deviation?

Group A: 11, 12, 13, 14, 15

Group B: 50, 51, 51, 52, 53

Even though the numbers in Group B are larger than those in Group A, they’re closer together thanks to the double occurrence of number 51. No such overlapping occurs in Group A. Group A exhibits a slightly greater spread and therefore has the greater standard deviation.
The 34-14-2 Rule
Standard deviation also lets you know how likely it is that a value will differ from the mean by a certain amount. In general, the farther a value is from the mean, the less likely it is to occur. The following graph, called a Normal Distribution, shows this in more detail:
This graph is the basis for the 34-14-2 Rule:
34%, 14%, and 2% represent the likelihood that a value will fall into each given region. For example, there is a 34% chance that a value will be between the mean and one standard deviation to the right or left of the mean. Similarly, there is a 14% chance that a value will be between one and two standard deviations to the right or left of the mean.
Here’s how to use the numbers. Say that the mean for some group of values is 10, and the standard deviation is 2. One standard deviation to the right of the mean will therefore be 10 + 2 = 12. The Normal Distribution graph states that there is a 34% chance that a value from the group will fall between 10 (the mean) and 12 (one standard deviation up from the mean). Based on this kind of analysis, you may be asked something like this:

What is the likelihood that a value within a group of values with a mean of 10 and a standard deviation of 2 equals 5?

If the mean is 10, then one standard deviation below the mean is 10 – 2 = 8, which creates a 34% chance that a value from the group falls between 10 and 8. A second standard deviation to the left would be 8 – 2 = 6, meaning that there’s a 14% chance that a particular value would fall between 8 and 6. More than two standard deviations to the left of the mean would be all values below 6. The graph tells us that these values have a 2% likelihood. The number 5 falls into this group, so 2% would be the answer.
Frequency Distribution
Say, for example, that fifteen college graduates were asked how many different jobs they had in their first five years following college, and the responses came back like so:
2, 4, 2, 1, 0, 1, 3, 2, 2, 5, 3, 4, 1, 2, 1
Not very pleasing to the eye, is it? One way to organize this information is to express it in the form of a frequency distribution: a chart that shows at a glance all of the answers given and the number of people who gave each answer. Frequency distributions typically designate x as the values (in this case, the answers given by those surveyed) and f as the frequency of each value. In the example above, that would look like this:
x f
0 1
1 4
2 5
3 2
4 2
5 1
Total 15
We can see from the chart that one person surveyed had no jobs in the first five years after college (slacker!), four people had one job, five people had two jobs, and so on. So what’s so great about this? Well, the best thing about it is that it allows us to quickly determine many of the other statistical features we’ve been discussing so far. For example, eyeing only the left-hand column tells us that the range of responses is 5 – 0 = 5. A quick scan of the right-hand column indicates that 5 is the largest frequency corresponding to any one answer, and it corresponds to the answer 2, which therefore qualifies 2 as the mode. The chart already lists the responses in ascending order, so the median will be the eighth value from the beginning—eighth because with fifteen values total, the eighth value is right in the middle with seven values below it and seven above it. The first value is 0, the next four values are 1, bringing us to the fifth value, and the next five values are 2, bringing us to the tenth value. The eighth value is therefore a 2, so 2 is the median of this group of values.
You may also be asked to calculate the mean from a frequency distribution. Recall the formula for mean:
Using the frequency distribution, we can quickly calculate the sum of the terms by finding the sum of each term multiplied by its frequency. The number of terms will be either given in the table (such as “total = 15” in our chart), or you can just add up the frequency numbers in the right-hand column to calculate the number of terms. In this case, the mean is:
If you learn the basics of frequency distributions, it should be a welcome sight if one of these appears on your test.
Probability
Probability is the measure of how often something is expected to occur, expressed as a fraction or decimal between 0 and 1. A probability of 0 means there’s no chance that the event under consideration will take place. A probability of 1 means it definitely will happen. Most probabilities tested on the GRE fall somewhere in between. We’ll use the common scenario of selecting colored marbles from a bag to illustrate the various kinds of probability questions you might see on your test.
Single Trials
The most basic kind of probability question involves a single selection from a given group of elements. Here’s an example:

In a bag containing 12 red, 13 white, and 15 black marbles, what is the probability of selecting a red marble on a single draw?

To tackle probability problems, use the following formula:
The number of favorable outcomes is math lingo for the number of ways you can get what the problem is asking you to get. Here, red marble is the favorable outcome, so the numerator of the fraction is 12, the number of red marbles in the bag. The total number of possible outcomes is the total number of possibilities, or, in our problem, the total number of marbles. Make sure to include all of the marbles, including those already counted as favorable outcomes. The total number of marbles is 12 + 13 + 15 = 40. Plugging 12 and 40 into the formula gives:
Simplifying this gives , the final answer.
Independent Events
If one event does not influence the occurrence or nonoccurrence of another event, the two events are independent. To find the probability of two independent events occurring, simply multiply their individual probabilities. For example, if there’s a 1 in 4 chance that Mary will be selected for a committee, and a 1 in 3 chance that Bill will be kicked out of college, and the events are independent (that is, Mary isn’t angling to join the committee with the purpose in mind of booting Bill), then there’s a chance that both Mary will be selected for the committee and Bill will be given his college walking papers.
We mention independent events at this point because this concept affects our next topic, multiple trials.
Multiple Trials
Frequently, probability questions on the GRE won’t be limited to a single draw, or trial, but will instead involve repeated draws. When a question involves drawing multiple times from the same group of entities, you need to distinguish between draws with replacement and draws without replacement. Let’s illustrate the difference using our marble example:
  1. 1 You select a marble, note its color, and put it back in the bag. You then select a marble again. This is called drawing with replacement.
  2. You select a marble and put it aside. Then you draw another marble from those remaining. This is called drawing without replacement.
The GRE will always make it clear which method is being used either by including the actual phrase with replacement or without replacement or by explicitly describing the method of selection in a way that makes it obvious which mechanism is in play. Let’s look at an example of each type.
Drawing with Replacement
Try your hand at this one:

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting two black marbles in a row if the selection is made with replacement?

The number of black marbles, or favorable outcomes, is 15. The total number of marbles is 40. First, use the probability formula to find the probability of selecting a black marble on the first draw:
Since this problem involves drawing with replacement, we’ll need to put the black marble selected on the first draw back into the bag before selecting again. So the bag will still contain 15 black marbles out of 40 total for the second draw. The probability of drawing a black marble on the second draw is thus the same .
Now, even though the marbles are coming from the same bag, these two events—a black marble on the first draw and a black marble on the second—are independent; that is, what happens on one draw doesn’t affect what happens on the other. To get the probability of two black marbles in a row, we can therefore multiply the individual probabilities:
Drawing without Replacement
Now let’s see what happens when we don’t put the first marble back into the bag after selecting it:

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting two black marbles in a row if the selection is made without replacement?

The probability of the first marble being black is , just as before. For the second draw, however, only 14 black marbles remain out of 39 total. (Remember, we took a black marble out of the bag and did not put it back.) This means that the probability of the second marble being black is . By assuming the first draw was favorable (a black marble selected), we adjusted the figures for our second probability. Since these figures are already adjusted to account for the first favorable outcome, the second drawing is independent from the first, so we can still multiply the individual probabilities to get the chances of selecting two black marbles in a row:
We can cancel the 3 and 39 before multiplying and also cancel a factor of 2 from the 8 and 14 to make our lives easier:
This can’t be reduced any further, so it’s the final answer.
The Probability of Something NOT Happening
You’re probably familiar with the phrase not happening, as in when you ask your boss for three weeks off in the summer and he tells you “that is so not happening.” But do you know that there’s an actual math formula for “not happening”? You probably do, without even realizing it. For example, if you’re told that the chance of snow tomorrow is 25%, it’s likely you recognize without much thought that the chance that it will not snow is 75%. Here’s the formula you used, whether you were aware of it or not:
the probability of an event NOT happening = 1 – the probability of that event
This formula can turn very hard probability questions into easier ones. Consider this next one:

A bag contains 12 red, 13 white, and 15 black marbles. What is the probability of selecting at least one red or one white marble in two draws if the selection is made with replacement?

This is harder than the previous problems, because it’s not altogether clear what must happen on any individual draw for a favorable outcome. For instance, the first draw might be black, and you still could have a favorable outcome if the second draw is red or white. Similarly, the second draw could be black, and you’d still have a favorable outcome if the first draw is red or white. And of course, a first and a second draw of red or white would also count as a favorable outcome. So how do we deal with this ambiguity?
Simple: Use the formula for “NOT happening.” It’s far easier in this case to calculate the probability of not getting at least one red or white marble in two draws because this is actually the same thing as drawing two black marbles, with replacement. We already calculated this earlier as . The probability of drawing at least one red or white in two draws is 1 minus the probability of that NOT happening, which is simply the probability of drawing two black marbles. The answer is therefore:
Still no piece of cake, but doable.
“Or” Questions
A difficult question may ask you for the probability of event A or event B occurring, which is different from the probability of A and B occurring. If you see one of these, use the formula:
probability of A or B = probability of A + probability of B – probability of A and B
For example, say the probability of Marcie passing a test is 70%, and the probability of Jerome passing the same test is 30%. The events in this problem are independent: Neither person passing the test influences whether the other person does so (unless of course they cheat from each other, which we’ll assume is not the case). So the final term of the expression, the probability of Marcie and Jerome passing, is equal to the product of the individual probabilities of those events, as we’ve seen all along. We’ll convert the percentages to fractions, since working with those may be easier. Then we’ll plug ‘em into the formula:
The probability of Marcie or Jerome passing the test is therefore equal to 79%.
In some cases, two events may be mutually exclusive, meaning that the probability of both occurring is 0. For example, in choosing a single dog from a kennel, the chances of choosing a black Labrador and a white schnauzer are zero—you can’t have both. If these represented the A and B elements of the “or” formula, the final term would be 0, and you wouldn’t have to subtract anything.
Sequences
A sequence is a list of numbers that follows a particular pattern. If you get a sequence problem, you’ll probably be given at least one of the terms in the sequence, along with the rule that defines the pattern. You probably won’t have to figure out the pattern on your own; that’s more like the kind of thing you’d see on an IQ test.
However, what could make sequence problems tough is the notation. Each term in a sequence has the same variable, but each has a different subscript. This subscript indicates a particular term. For example:
a1 = the first term
a2 = the second term
a10 = the tenth term
a n = the nth term
an+1 = the term immediately after the nth term
For example, to indicate that the second term of a sequence is 5 and that the third term is 7, the test makers might write:
a2 = 5
a3 = 7
This subscript notation can also be used to indicate how each term relates to the others. For example:
an+1 = an – 3
This just means that each successive term is three less than the previous term. Here’s an example, using the notation we just discussed:

If a n = the nth term in a sequence, and a1 = 3 and an+1 = a n + 2, what is the value of a10?

Let’s use 1 as n to keep things simple:
a1 = a n = 3
So an+1 is the same as saying a1+1 or a2. This second term we’re told is equal to the first term, a n , plus 2, which means that the second term will be 3 + 2, or 5. So the notation, which looks intimidating, is really a shorthand way of saying that each successive term is two more than the previous term. Writing out this sequence from the first to the tenth terms gives 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. The tenth term is 21, and so a10 = 21.
Arithmetic Sequences
In an arithmetic sequence, the difference between each term and the next is constant. This is the kind of sequence we saw in the previous example. In addition to understanding the notation and concepts for sequences, you should know the formula for arithmetic sequences:
a n = a1 + (n – 1)d
where
a n = the nth term
a1 = the first term
d = the difference between consecutive terms
This formula is useful if you need to determine the value of some very high term and don’t want to write down a long sequence of numbers. In our previous example, the first term (a1) is 3 and the difference between consecutive terms is 2. If you plug these numbers into the equation above looking for a10, you’ll get the same answer, 21, that we got earlier. In this example, it’s just as easy to write out the terms. But to determine the 100th term in that sequence, we’ll need to plug the numbers into the formula:
a100 = 3 + (100 – 1)2 = 3 + 99 × 2 = 201
Geometric Sequences and Exponential Growth
In a geometric sequence, the ratio between one term and the next is constant, not the actual difference between the terms. For example, in the sequence 3, 9, 27, 81, each successive term is three times greater than the preceding one, but the actual difference between the terms changes: 9 – 3 = 6, 27 – 9 = 18, and so on. Geometric sequences exhibit exponential growth, as opposed to the constant growth of arithmetic sequences. Here’s an example of the kind of geometric sequence that the test makers might toss at you:
g1 = 4
g n = 2gn – 1
Trying out some terms, this means that g2 = (2)g1, g3 = (2)g2, and so on. In other words, the first term is 4, and each successive term is twice the value of the preceding term. Writing out the first few terms lets us see that the ratio between terms is constant and thus confirms that this is a geometric sequence:
4, 8, 16, 32, . . .
The ratio between consecutive terms is always 2, even though the differences between the terms increase as you move to the right.
As with arithmetic sequences, you should learn the special formula for geometric sequences, just in case it’s not convenient to list out all of the terms up to the one you’re looking for:
g n = g1rn–1
where
g n = the nth term
g1 = the first term
r = the ratio between consecutive terms
Let’s use the formula to calculate the value of the tenth term in the geometric sequence defined by g1 = 4 and g n = 2gn – 1. We already know r = 2:
g10 = 4 × (2)10–1 = 4 × 29 = 4 × 512 = 2,048
Digit Counting
Here’s an interesting kind of problem that appears with some regularity on the GRE. In digit counting problems, you’re asked how many times a particular digit appears in a defined group of numbers, or how many numbers within such a group don’t contain a certain digit. An example will make this clearer:
How many three-digit positive even integers contain at least one digit that is a 7?
(A) 45
(B) 60
(C) 90
(D) 140
(E) 210
First make sure you understand the range of numbers under consideration. Positive three-digit numbers begin with 100 and go to 999. Moreover, we’re only interested in the even ones. The question is looking for how many numbers fit this description and contain at least one 7.
Well, we’re not going to go and count them all—that would take too long. But we will list a few examples that fit the criteria and then see if we can discern a pattern. Beginning with the 100s, the first number that has a 7 is 107, but since that’s not an even number, it doesn’t count. In fact, the next one, 117, doesn’t count either, nor does 127, 137, and so on. So we begin to see a pattern. The first even number we get to that has a 7 is 170, followed by 172, 174, 176, and 178. No number from 179 to 200 fits the bill, so in the entire 100s we have a total of 5 numbers that satisfy the question’s requirements.
Now we can generalize from what we’ve learned: The 200s will be no different, nor the 300s, nor the 400s, and so on. So if there are 5 cases in each of the nine groups of 100 numbers from 100 to 999, we can multiply 5 instances per each group of hundreds by nine groups of hundreds (100s, 200s, etc.) to get 45, choice A.
However, we’d be wrong. Naturally, A is a trap for people who forget that the 700s contain plenty of numbers that work, so we need to consider the numbers from 700 to 800 separately. The other eight groups of hundreds follow our pattern, so we’ll go with 5 × 8 = 40 to represent the instances that make the cut among those. But we also have to consider all of the numbers in the 700s, since every number in the 700s contains at least one 7. How many are even? There are 100 numbers between 700 and 799, and half are even, so we need to add 50 more cases to the 40 we’ve found already. The correct answer is 40 + 50 = 90, choice C.
Digit counting questions can be difficult, and if you see one, chances are the GRE software is throwing you more difficult questions because you’re doing pretty well. Understand the range, find a pattern and generalize it to as many cases as you can, and then check for special cases. If you’re careful, you’ll find them all.
Factorials
You may see some problems on your exam in which a number is followed by an exclamation point, like this: 5!. This does not mean you should loudly exclaim, “Five!” Nor does it mean that the test makers are extra-specially enthusiastic about a particular problem. The exclamation point makes it look like there’s something really exciting going on! But there’s not.
An exclamation point used in math symbolizes a factorial. A factorial stands for the product of all the numbers up to and including the given number. So 5! = 5 × 4 × 3 × 2 × 1 = 120.
Some more examples:
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
55! = 55 × 54 × 53 × . . . × 3 × 2 × 1 = a really huge number you would never be expected to solve for
0! = 1
The proof of this last example is beyond the scope of what you need to know for the GRE. Just remember that 0! = 1 by definition. Consider it another bit of math trivia picked up on your way to GRE mastery.
The factorial of n also signifies the number of ways that the n elements of a group can be ordered. So, if you decide to ditch grad school and become a wedding planner instead, and need to figure out how many different ways six people can sit at a table with six chairs, 6! is the way to go: 6 × 5 × 4 × 3 × 2 × 1 = 720 possible seating arrangements. You’ll astound the other wedding planners with this quick calculation, never revealing the true source of your knowledge.
As you might guess from the name, factorials have many factors. Recall that a factor is a number that divides into another number with no remainder. Whenever you take the factorial of a number, the result will be divisible by all of the integers up to and including the original number. For example, 6! is divisible by 6, 5, 4, 3, 2, and 1, and all of those numbers are factors of 6!. This is all inherent in the definition of factorial, but it’s good to understand it in these terms too.
Simplifying Factorials
The test makers may ask you to work out a problem that involves factorials in fractions, and as you’ll soon see, this becomes downright necessary in permutation and combination problems. The trick is to cancel before calculating. As you’ve seen in earlier examples, canceling with fractions means dividing the numerator and denominator by the same number. A little cancellation makes complicated-looking factorial problems much easier to solve. Check it out:
What is ?
This expression looks like it might be a huge number. And, in fact, trying to calculate 98! or 100! would be near impossible without a computer or ultra-fancy calculator. Fortunately, we can simplify this equation significantly:
This works, because everything after and including the 98 cancels out in both the numerator and the denominator, leaving 100 × 99 in the numerator and 1 in the denominator. Here’s another way to think about this:
The 98! in the numerator cancels out with the 98! in the denominator, leaving only 100 × 99 = 9,900.
Truth be told, even factorial problems involving smaller numbers benefit from canceling out. For example:
Before you get sucked into multiplying out the factorials in the top and bottom parts of the fraction and then dividing the results, first cancel out what you can.
Permutations and Combinations
Knowing how to calculate and simplify factorials is especially useful for problems involving permutations and combinations. These types of questions ask you to determine how many ways something can be done. They’re similar to the simple wedding planner example above, except that they involve choosing a certain number of entities from a larger group. For example, “In a race of eight horses, how many ways can the horses finish first, second, and third?” and “How many ways can two students be selected for a Grammar Jamboree out of a class of 20 students?” In this section, we’ll explain not only how to answer both of these questions but also the very important difference between them.
Permutations
In a permutation, order matters—that is, being first in a group is different from being second, third, or in any other position. The easiest way to tell that a question is a permutation is if it includes the word order or the word arrange. Even if it doesn’t contain these words, the question might describe some kind of ranking or race. Our horse race question above, for instance, is a permutation since finishing first is certainly different from finishing second or third in a horse race. If, for example, three of the horses are A, B, and C, then ABC is one possible finish, CAB is another, and BCA is a third. They all involve the same three horses, but switching them around yields additional arrangements that we need to add to our tally. In a combination problem, however, we’re not concerned with order, so BCA would be considered the same as CAB, and we wouldn’t count those twice. We’ll get to combinations in just a bit, but let’s continue with the permutation problem at hand.
Here’s the permutation formula:
In this case, nPr is the number of subgroups of size r that can be taken from within a set with n elements.
Here’s the horse race question again, and this time we’ll work through it using the formula:

In a race of eight horses, how many ways can the horses finish first, second, and third?

In this problem, n = 8, because that’s the total number of horses racing, and r = 3, because that’s the number of winners we’re interested in (first, second, and third place). Plugging into the formula gives:
Now we’ll solve, using our knowledge of factorials and canceling out:
Notice how the 5!s cancel out, leaving us with some basic multiplication to get our final answer.
Combinations
As we mentioned earlier, in a combination, order does not matter. For example, if you’re trying to buy three horses instead of ordering them first through third in a race, then it really doesn’t matter if you come away from the horse farm with horses ABC or horses CBA—they’re all the same horses. You wouldn’t shuffle them around and then say, “Look, a whole new group of horses! Lucky me!” Well, you could say that, but people would think you’re nuts.
You’ll be able to recognize a combination problem because it will involve selecting a small group from a larger group, with no regard to order. An example is the Grammar Jamboree problem introduced earlier:

How many ways can two students be selected for a Grammar Jamboree out of a class of 20 students?

Two students are to be selected from a class of 20, and no mention of order is made. This is common for situations involving teams: A team consisting of Jonathan and Gloria is the same as a team consisting of Gloria and Jonathan. Unless some specific mention is made of an ordering element, assume you’re dealing with a combination problem. Here’s the combination formula:
Here, unordered subgroups of size r are selected from a set of size n. Notice that this is the same as the permutation formula, except that it tacks on an extra r! term in the denominator. This means that we divide by a larger number in combination problems, resulting in a smaller number of final orderings. And that makes sense too: We’d expect fewer total orderings in combinations, since order doesn’t matter and we therefore count the shuffled orderings (ABC, CAB, BCA, etc.) as one.
Use the formula to solve our Grammar Jamboree problem:
Cancel the 18!s:
So, there are 190 possible two-person teams of jamboree-ers to choose from the class of 20. Let the jamboree begin!
Multiple Permutations and Combinations
If the CAT software is really impressed by your math acumen, it might throw you a problem involving multiple permutations, combinations, or both. The key is to break the problem down into parts, solving each independently using the formulas discussed above. To obtain a final answer, multiply all of the individual results. Here’s an example:

How many ways can Suzanne order an ice cream cone if she is to select three different flavors out of fifteen available flavors and three different toppings out of five available toppings? The order in which the flavors are stacked is significant, but the order in which the toppings are added is not.

We admit it—this looks hard. However, it helps to think of this as two completely separate problems:
  1. Selecting ice cream flavors
  2. Selecting toppings
Don’t worry, we’ll combine them in the end, much as the flavors and toppings will be combined in Suzanne’s cone. But first things first.
Since the order of flavors matters, this part of the problem is a permutation. Suzanne must select three different flavors out of a total of fifteen. Let’s use the P formula:
As per the problem, n = 15 and r = 3:
So Suzanne has 2,730 choices of flavors. Delicious! But back to the problem: Since the order of toppings does not matter, this part is a combination. Suzanne must select three different toppings out of a total of five. Let’s use the C formula:
Now n = 5 and r = 3:
Cancel the 3!s, and restate 2! as 2 × 1:
Now we know that there are 10 possible combinations of toppings. The last step is to multiply the two results: 2,730 × 10 = 27,300. And there we go: Suzanne has an unbelievable 27,300 choices for her ice cream cone. It’s a wonder people don’t have anxiety attacks in ice cream stores—just one cone but more than 27,000 varieties to choose from!
Groups
This topic sounds innocuous enough, but problems involving groups can drive test takers batty. The basic idea is that some people or things belong to one group, others belong to another group, and still others belong to both groups or neither group. For example, at a certain country club, some members play golf, some play tennis, others play both sports, and still others prefer reading to playing. You’ll be given some of the specific numbers in such a problem and then asked to determine the missing values.
The approach you should use depends on whether the problem concerns two or three groups. We’ll cover the most effective techniques for both cases. You won’t see problems with more than three groups.
Problems with Two Groups
All you need for two-group problems is this formula:
group 1 + group 2 – both + neither = total
where
group 1 = the number of entities in one of the two groups
group 2 = the number of entities in the other group
both = the number of entities in both groups
neither = the number of entities in neither group
total = the total number of entities
You’ll most likely be given values for all of the parts of this formula except for one. You’ll then have to determine the value of the missing part. Let’s see how this works in the following example.

At a certain animal refuge, 180 animals have four legs, 240 are warm-blooded, and 85 both have four legs and are warm-blooded. If the animal refuge has 500 animals in total, how many animals at the refuge have neither four legs nor are warm-blooded?

If you don’t know the formula above, this question could be a nightmare. But the formula makes it very doable. We’ll let group 1 be the 180 animals that have four legs and group 2 be the 240 animals that are warm-blooded. Since we’re also given the number of animals that belong to both groups and the total number of animals, the missing value is the number of animals that belong to neither group. Not surprisingly, that’s what the question is after. Plugging the values into the formula gives:
180 + 240 – 85 + neither = 500
Solving for neither is a simple matter of solving this linear equation with one variable, something we discussed way back in the algebra section. Here goes:
180 + 240 – 85 + n = 500
335 + n = 500
n = 165
Voila! (That’s genuine excitement—not a factorial.) But things get a bit more difficult if they throw in an extra group.
Problems with Three Groups
The formula for three-group problems is really long and complicated (it involves nine unique terms!). So, we’ll skip it in favor of an easier approach: Venn diagrams. Remember these from junior high? Venn diagrams consist of intersecting circles, in which each circle represents the number of entities in a particular group. For example:
You’ll notice that the circles overlap. The upside-down triangular section in the middle, with the darker shading, represents the number of entities that belongs to all three groups. Sections in which only two circles overlap, indicated with the lighter shading, represent the number of entities that belongs to two overlapping groups. The outermost section of each circle, the part that doesn’t overlap with any of the other circles, represents the number of entities that belongs to each group alone. For example, in the swimming circle, the outermost section represents the number of swimmers who neither lift weights nor do aerobics.
The key to three-group problems is to work from the inside out. Begin with the entities that belong to all three groups, then address the entities that belong to two groups, and finally deal with the entities that belong to only one group.
Here’s an example of a three-group problem that conveniently makes use of the diagram above:

At the Get Fit Athletic Club, every member swims, lifts weights, does aerobics, or participates in some combination of these three activities. Sixty members swim, 75 lift weights, and 100 do aerobics. If 34 members both lift weights and swim, 25 members both do aerobics and swim, 44 members both use the weight room and do aerobics, and 10 participate in all three activities, how many members belong to the Get Fit Athletic Club?

We’ll start by filling in values, working from the inside out. Since 10 people belong to all three groups, we’ll write 10 in the middle section:
Next, we’ll fill in the values for people who belong to two groups. We’re told that 25 members participate in aerobics and in swimming, and it would be very tempting to write 25 in the section above the 10. Keep in mind, however, that 10 of these 25 people have already been accounted for: The 10 people who participate in all three activities are included among those who participate in aerobics and swimming. That leaves 25 – 10 = 15 people for the aerobics/swimming overlap section above the 10 in the middle. Similarly, 44 people do aerobics and use the weight room. Since 10 of these people have already been accounted for in the middle section, that leaves 44 – 10 = 34 people in the section that overlaps aerobics and weight room. Also, 34 people swim and use the weight room, so we’ll write 34 – 10 = 24 in the section overlapping those categories. That brings us to here:
Next we need to fill in values for people who belong to only one group. As with the previous step, we have to be very careful not to count anyone more than once. For example, we’re told that 60 members swim. That means that the total of all the numbers in the swimming circle must be 60. We already have 10 + 15 + 24 = 49 members in the swimming circle. That leaves 60 – 49 = 11 members for the outermost section of the swimming circle. Similarly, since 75 members use the weight room, that leaves 75 – 10 – 34 – 24 = 7 members for the outermost section of the weight room circle. Finally, since the total of the aerobics circle must be 100, the number in the remaining section must be 100 – 10 – 34 – 15 = 41. Filling these numbers in the appropriate sections of our Venn diagram yields the following:
The problem asks for the total number of club members, and we’re told specifically that every member participates in at least one of these activities. That means that no member exists outside of our three circles. We can therefore add all of the values in our diagram to arrive at the total number of club members. This gives us a final answer of 41 + 15 + 11 + 34 + 10 + 24 + 7 = 142.
We realize that this was a bit of work, but Venn diagrams are the only way to go on three-group problems. And the good news is that you’d have to be doing something right on the Math section for the CAT software to throw something this nasty your way, since the questions increase in difficulty the better you do.
That finishes up our discussion of data analysis. Keep in mind that the GRE also requires you to analyze data presented in the form of graphs and charts, but we have a whole chapter called Data Interpretation devoted to that. In fact, we have a special chapter devoted specifically to each of the three question types you’ll see. It’s high time we got to those, where you’ll learn how the concepts covered here in Math 101 play out in GRE math questions, as well as effective ways to go about tackling them.
Problem Solving is the first question type we’ll cover, and it’s up next.
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