GRE Math: Factorials, Probability, Arrangements, and Combinations, Study Guides, Projects, Research of Mathematics

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Chapter 9

The Rest of the Story

TYING UP LOOSE ENDS

So here it is, the last chapter about quantitative topics on the GRE. Most of the material we review in this chapter will probably appear less frequently on the GRE, and most of it deals with data analysis. There are a lot of formulas in this chapter that you’d do well to memorize, because it will save you a considerable amount of time when you sit down to take the test. Let’s begin.

FUNCTIONS

Sometimes the GRE will make up a mathematical operator. You will see some weird shape, which you’ve never seen in a math problem, and be asked to solve a problem using this weird shape. This is simply the GRE trying to confuse you.

These questions are asking about functions. A function is simply a set of directions. For instance, think of the word “chop” in a recipe. The word chop is actually telling you to do many things: take out a cutting board, rinse the vegetable or fruit to be chopped, take out a knife, place the vegetable on the cutting board, etc. If a recipe says “Chop 3 carrots,” then you must do all those things that chopping entails, but using carrots. If the recipe says “Chop 2 stalks of celery,” then you must do all of those chopping things, but with celery.

A function is the same way. It is a set of rules, and the GRE will ask you to perform those rules on a certain number. If the GRE invents some function, for instance that ♦ x = 4 x + 2, and then asks what ♦5 is, then figure out what rules you need to follow. The original function was ♦ x , but notice how ♦5 replaced the x that was after the ♦ with a 5? Well, do the same thing with the equation given for ♦ x : Replace each x with a 5. You get ♦5 = 4(5) + 2 = 20 + 2 = 22.

Try a practice problem. Look for what numbers you’ll need to place in the original function, and where each number will go.

Here’s How to Crack It Here you have a completely made-up mathematical operator. This question says that whenever you see a ~, you must find the sum of all the prime integers between those two numbers. Rather than write out all the primes and figure out which numbers you could pick to get 17, try PITA. As you plug in each answer choice, find the sum of all the prime integers between those two numbers. So 1~10 is the sum of all the prime integers between 1 and 10, which is 2 + 3 + 5 + 7 = 17, so (A) is an answer. Since 4~10 = 5 + 7 = 12, cross off (B). Since 6~12 = 7 + 11 = 18, cross off (C). For the last answer, the only prime integer between 14 and 18 is 17, and since the sum of 17 and nothing is 17, (D) is also an answer. The answers are therefore (A) and (D).

Here’s How to Crack It

15! is too large to enter into the calculator, so factor out what 15! and 14! have in common. Since 15! is the same as

15 × 14 × 13 × 12 × 11 × 10…and 14! is 14 × 13 × 12 × 11…, rewrite 15! as 15(14!). Both 15(14!) and 14! contain

14!, so rewrite the fraction as. Note that if you distributed the 14! to each term within the parentheses,

you’d have back the original numerator: 15! − 14!. You can simplify the 15 − 1 inside the parentheses to get.

Now it’s time to cancel out the two factorials, as you did with the previous example question. =

Factorials Quick Quiz

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Explanations for Factorials Quick Quiz

1. The first term, , is equivalent to 25 × 24, and the second term equals. Therefore, the final term is

, or 25.

2. All of the answer choices are equivalent to 6!, or 720, except (E), because is the same as . Because 12 × 11 × 10 is already 1,320, it’s possible to see

that this is far bigger than 720 without even bothering with your calculator. The answer is (E).

3. Since there are no variables in this question, eliminate (D). There are factorials combined with subtraction, so you’re going to have to factor. 17! is the same as 17 × 16 × 15 × 14!. You can therefore rewrite Quantity A as 17 × 16 × 15 × 14! − 14!. Since you have 14! in both terms, factor out 14! to get 14!(17 × 16 × 15 − 1). Now it’s time for a bit of calculator work, which gives you 14!(4,080 − 1) = 14!(4,079). Both quantities have 14!, so focus on the other parts. Since 4,079 is larger than 4,078, 14!(4,079) is larger than 4,078 × 14!, and the answer is (A).
4. There are variables in the answer choices and question stem, so Plug In. Pick a number that is easy to work

with for n , such as n = 5. If n = 5, then the target answer can be found by reducing the expression in the

question stem. The question stem now reads which, if expanded, reduces to 6 × 5 × 4 = 120, which is

the target answer. Now, plug in 5 for all values of n in the answer choices and see which results in a value

of 120. Choice (A) is clearly incorrect, so eliminate (A). Choice (B) is 5!, which expands to 5 × 4 × 3 × 2

× 1 = 120, so keep (B). Choice (C) is 5^3 − 5 = 125 − 5 = 120, which is also correct so keep (C). Choice

(D) is less than (C) because 4(5) = 20, and 125 − 20 = 105, so eliminate (D). Choice (E) is 10!, which is

too great, so eliminate (E). Now Plug In again for (B) and (C), but this time use a FROZEN number. Try

an Extreme number such as n = 10. If n = 10, then the question stem now reads , which, if expanded,

reduces to 11 × 10 × 9, which equals 990. With 990 as the new target answer, plug in 10 for (B) and (C) to

Multiple Probabilities

If you’re asked to find the probability that two specific events will occur one after the other, first find the individual probabilities that each event will occur, and then multiply them. There are two different scenarios on problems like these; sometimes the odds of an individual event are different from another event, and sometimes they aren’t.

Probability of A and B = Probability of A × Probability of B

For example, if a problem involves flipping a coin, the probability that it will come up heads will always be. It will

never change, because there will always be heads on one side of the coin and tails on the other. Here’s a typical

problem you might see on the GRE.

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Here’s How to Crack It

There is only one 5 on the six-sided die, so the chance of rolling a 5 is. When you roll the die a second or third time,

you still have a chance of rolling a 5, so the chance remains. Therefore, the chance of rolling three 5’s in a row is

. The answer is (E).

Removing Items Changes the Probability

So far, the total, the denominator of our fraction, has always stayed constant. However, there will be some questions in which items are removed as you go. These problems often contain the phrase without replacement , because items are taken without replacing them. In that case, the total will change. At each point, figure out the next probability by assuming that what the problem wants to happen has happened.

For instance, imagine there are 3 red marbles and 1 green marble in a bag. What’s the probability of selecting 2 red

marbles in a row? When you first reach into the bag, you’ll have a chance of pulling out a red marble: There are 3

red marbles, out of a total of 4 marbles. But what about when you reach in again? You’ll have to assume that you

pulled out that first red marble. (Otherwise, who cares if you pull out a second red marble? You wanted two red

marbles in a row, not a something else and then a red marble.) That means you have only 2 red marbles left, out of a

total of 3 marbles, and your chance to pull that second red marble is. Now you can combine your two probabilities

by multiplying them together: × ×.

Try a harder problem that uses a changing total.

The probability that the first two peppers are green and the third is red

The probability that the first pepper is green and the last two are red ○ ○ ○ ○

Here’s How to Crack It

This question is different because multiple peppers are removed, so the odds change after each pepper leaves the bag.

Work with Quantity A first: When the grocer reaches in the first time, he has a chance of selecting a green pepper.

The second time, there are only 4 green peppers out of a total of 8, so the chance of getting a second green pepper is

. The third time, there are only 7 peppers left and 4 of them are red. The chance of getting a red is , and the

composite probability is.

Before you multiply, look at Quantity B. The probabilities are (green), (first red), and (second red), so the

composite probability is. When comparing the two quantities, the denominators are the same but the

numerator in Quantity B is smaller. Therefore, Quantity B itself must be smaller, and the answer must be (A).

At Least Probabilities

Some questions will not ask for the probability of a single event, but instead ask for the probability that “at least” a certain number of events will happen. For instance, let’s go back to the bag of marbles. Say there are 4 red marbles and 5 green marbles, and you’re asked the likelihood of reaching in, taking out 3 marbles, and getting at least 1 red marble.

You could find that out by figuring out the probability of getting 1 red marble and then 2 green marbles, and adding that to the probability of getting a green, a red, and a green, plus the probability of getting a green, a green, and a red. But you could get more than just one red, right? You’d also have to find the chance of RRG, RGR, and GRR, and then the probability of getting RRR, and add all of those different ways of getting at least one red marble together.

But there’s an easier way. Think of it this way: If there’s a 30% chance of rain, what’s the chance it won’t rain? 70%. Since any two mutually exclusive probabilities always add up to 100%, you can sometimes find the probability that something won’t happen rather than find the probability it will happen. Once you do, subtract it from 1, because

The probability that he will roll numbers that sum to 8 The probability that he will roll numbers that sum to 7 ○ ○ ○ ○

Explanations for Probability Quick Quiz

1. There are 50 different states; these are the possible outcomes. There are 23 states with a saltwater border,

and these are the desired outcomes for Quantity A. Therefore, the probability that a randomly selected

state will have a saltwater border is. If 23 have saltwater borders, then 27 states do not. This is the

desired outcome for Quantity B. The number of possible outcomes remains the same, so the probability

for Quantity B is. This fraction is greater, so the correct answer is (B).

2. There are 24 (or 4!) ways to randomly select the numbers 1, 2, 3, and 4. There is only 1 way to select them

in ascending numerical order, so the probability is. The correct answer is (C).

3. There are 9 coins in Fred’s pocket, 3 of which are dimes, so the probability that he will get a dime the first

time is. The second time, there are only 2 dimes out of 8 coins, so the probability drops to. The

composite probability is × , or. This reduces to , so the correct answer is (A).

4. When two fair, six-sided dice are thrown, there are 36 possible rolls. Of these, 5 will sum to 8 (2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2) and 6 will sum to 7 (1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, and 6 and 1). Because there are more ways to roll 7, Quantity B is greater so the correct answer is (B).
5. Remember that the probability of two events happening is probability of A × probability of B. Since you

know that probability of J × probability of K = 0.42, and you know the probability of J is 0.63, you know

that 0.63 × probability of K = 0.42. Dividing both sides by 0.63 gives the probability of K:.

As we near the homestretch, here are two more topics that could appear on the GRE. These topics occur very rarely, and when they do appear you’ll see at most one question about them per section. Therefore, it pays to reference them here, at the end of the chapter.

GROUPS

There are two basic types of group questions: those that involve overlap between two groups, and those that involve groups in which there is no overlap. The group questions with overlap can be represented visually using the classic Venn diagram below, but you may find it easier, and more straightforward, to use an equation.

Group Formula: Total = [Group 1] + [Group 2] − [Both] + [Neither]

The first thing you’ll have to do with this sort of question is to recognize that you need to use the Group Formula. What are your clues? If there are any elements that overlap between two of the groups, then you will need to use the Group Formula. You may not have any elements that are in neither group (Neither = 0), but if you have any overlapping elements, then those elements are being counted twice: once because they belong to Group 1, and again because they belong to Group 2. To eliminate one of the times those Both elements are being counted, subtract the Both. Note that Group 1 includes everything in Group 1, including those that are in Both. The same applies to Group

Try a problem using the Group Formula:

Now take the question apart in bite-sized pieces. The first thing you find out is that there are 5 times as many students as faculty. If there are f total members of the faculty, then 5 f = s. Now that you know there’s a total of 12,000 faculty and students, 12,000 = f + s. Substituting in the earlier equation gives you 12,000 = f + 5 f. Since 12,000 = 6 f , f = 2,000. Since there are 2,000 faculty members, there are 10,000 students.

Now you can use the next piece of information: There are 200 more female faculty members than there are male faculty members. There are 2,000 total, so m + w = 2,000. You know that w = 200 + m , which you can plug into the earlier equation to get m + 200 + m = 2,000, so m = 900. If there are 900 male faculty, there are (2,000 − 900) 1, female faculty.

Okay, there is one more piece to use: There’s a total of 6,425 males. There’s a total of 12,000 students and faculty, which means there are 12,000 − 6,425 = 5,575 females. Fill that into the chart.

Now you can easily fill in the rest of the table with simple subtraction: The 6,450 − 900 = 5,550 males, and the number of female students is 5,575 − 1,100 = 4,475. The answer to the question is 4,475. (You don’t need to know the number of male students to answer the question, but for the sake of completeness, 6,425 − 900 = 5,525 male students.)

STANDARD DEVIATION

Standard deviation means almost exactly what it looks like it means: deviation from the “standard,” or mean, value of a set of numbers. A normal distribution of data means that most of the numbers in the data are close to the mean, while fewer values spread out toward the extremes. The bigger the deviation from the norm, the wider the spectrum of numbers involved.

The Bell Curve

A normal distribution is best displayed in the form of a regular bell curve, which looks like this:

The mean is the middle number, right at the 50% mark. The GRE will either just present you with the mean straight up, or it will be one quick calculation away. The rest of the lines on the curve represent standard breakpoints at 34%, 14%, and 2% of the data values. These mean that, within a normal distribution, 68% of the values (34% on the left, 34% on the right) are within one standard deviation from the mean.

To explain the normal distribution, use a sample data set. Say we asked 1,000 people to see how long they could hold their breath. After measuring all those people, and watching all those faces turn purple, we calculated that the average (arithmetic mean) number of seconds that people could hold their breath was 50. We then handed our data to a statistician friend, who calculated that our data followed a normal distribution and the standard deviation was 15 seconds.

Since the mean time was 50 seconds, half of the people were able to hold their breath for less than 50 seconds and half the people were able to hold their breath for longer. Our data would look something like this:

The same holds true in the other direction. 34% of people surveyed were able to hold their breath for anywhere between 35 and 50 seconds. Those people were 1 standard deviation or less from the mean. As we move another standard deviation of 15 seconds away, we find that only 14% of people could hold their breath for between 20 and 35 seconds. Finally, we have the bottom 2% of people. These people are 2 standard deviations or more below the mean, which is another way of saying that they are the absolute worst at holding their breath.

You will never have to calculate the standard deviation directly from the data on the GRE, but you will have to understand how it works.

• The bigger the standard deviation, the more spread apart the numbers. Imagine an unkempt field. If we went out and measured the height of the grass in that field, we’d probably get a standard deviation of about 20 centimeters or so: We’d have some really tall grass, but also some young, short grass. Due to those variations in height, we’d have a large standard deviation. The height of grass wouldn’t always be too close to the mean. Now imagine we took a lawnmower to that field. The lawnmower wouldn’t cut every blade of grass to the exact same height, but it would make every blade fairly close to the same height. Our standard deviation would shrink from 20 cm to 1 cm. Sure, some grass is a little taller than our mean height and some is a little shorter, but overall the heights of the blades of grass are very close to being the same. There aren’t many variations in height, so we have a small standard deviation. Lots of tall grass, short grass, and medium grass meant a large standard deviation, but when all of our grass was close to the same height, we had a small standard deviation.
• The more standard deviations away from the mean, the “stranger” you are. Say that the average pop song is three minutes long, with a standard deviation of 30 seconds. A song that is 3:30 isn’t that weird, because it’s only 1 standard deviation away from the mean. A song that’s 3:45 is a little more unusual, because it’s 1.5 standard deviations from the mean. A song that’s 4:30 long is really weird, because that’s 3 standard deviations away from the mean. For a pop song, it’s really unusual to be that long. Compare that to the opera. The average opera is 90 minutes long, with a standard deviation of 25 minutes. A two-hour ( minute) long opera, therefore, isn’t that weird: It’s a little more than 1 standard deviation away from the mean. But how about a five-minute opera? That’s around 3.5 standard deviations below the mean. For an

opera, that’s freakishly short.

That’s all you need to know about standard deviation. Sometimes, in fact, the GRE will even give you the 34-14- pattern and a drawing of the bell curve, but you should still memorize 34-14-2 and remember that the bigger the standard deviation, the more spread apart the numbers, and that the more standard deviations away from the mean, the “stranger” you are.

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Here’s How to Crack It If the average age is 61 and the standard deviation is 2.5 years, then a person who is 56 years old is 2 deviations from the mean (2 × 2.5 is 5, and 61 − 5 = 56). This means that only the bottom 2% of the group are less than 56 years old. The answer is (A).

Your scratch paper should look like this:

Each time we had a choice to make, we looked at the number of options we had for that choice. As we moved through our choices, we had fewer options, because we pretended that the previous choices had been made.

Try the next problem.

Here’s How to Crack It The end of the problem states in how many ways , which means this is an arrangement question. There are 3 ribbons to give away, so draw 3 slots on your scratch paper. For that blue ribbon, you could give it to whichever of those 8 horses comes in first, which means you’ve got 8 options for our first slot. Once you’ve given one of the horses a blue ribbon, there are only 7 horses left to win the red ribbon, so put a 7 in the second slot. Now there are 6 horses left to win the yellow ribbon, so put 6 in the third slot. You now have 8 × 7 × 6 = 336 different ways of awarding those ribbons.

If a problem contains restrictions on what you can choose at certain points, start with those restricted positions first, and then deal with everything else. It’s as though you were seating guests around a table. If you knew that two people just really, really, hated each other, then the first thing you’d do is make sure they sat on opposite ends of the table. Once that’s taken care of, you’d then place the rest of your friends in the remaining seats.

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Here’s How to Crack It The problem asks how many different lists , so this is another arrangement question. There are 5 songs you need to put in order, so draw 5 slots on your scratch paper. Here, however, you’re limited into what can go in the last spot in the list: It’s got to be one of 2 possible songs. So start by putting 2 in the last slot.

Now you can deal with the other slots. Since you have no other restrictions, continue by going back to the first slot. Since you put one of the songs in that last spot, you have only 4 songs left to put in the first slot:

The second slot therefore has only 3 songs possible, because you’ve already placed songs in the first and last slots.

Continuing to fill out the slots, you get

Multiplying 4 × 3 × 2 × 1 × 2 gives you 48, (B).

The questions we’ve done so far are all arrangement questions, because the order in which our elements are placed matters. Putting our Chico picture before our Harpo picture would be different from putting Harpo before Chico. Switching the first and second place ribbons of two of the horses would mean a different outcome of the race. Switching the middle songs of Boone’s playlist would mean a different playlist than we had originally. Arrangement questions are questions in which order matters.

What about when order doesn’t matter? Those are called combination questions, and they often contain words such as “groups,” “teams,” or (obviously), “combinations.” If you had to choose two books to loan, at the same time, to a friend, it wouldn’t matter if you loaned your friend Pale Fire and The Tin Drum as opposed to loaning her The Tin Drum and Pale Fire. Either way, she’s borrowing the same two books.

Since order doesn’t matter on combination questions, we’re going to add one extra step at the end. We need to get rid of the repeat groupings, so we’ll divide our answer by the factorial of our number of slots. An easier way to think of it is that we’ll count down to 1 underneath our slots.

The total number of different groups that could be chosen 120 ○ ○ ○ ○

Here’s How to Crack It Since there is a group of 4, draw 4 slots on your paper. You could choose one of the 9 people for the first slot, leaving 8 for the second slot, 7 for the third, and 6 for the fourth, giving you 9 × 8 × 7 × 6. However, you’re choosing groups of people, which means that the order you choose each group doesn’t matter. As long as it’s got the same people in it, who cares in what order those people were chosen?

The last step, therefore, is to divide by 4!. You can do this by counting down from 4 underneath each slot, making

each number a fraction. You now have , which you can simplify to = 126. Since Quantity

A is larger than Quantity B, the answer is (A).

By the way, did you notice how the denominator of the fraction canceled out completely? That will always happen with combinations. Think of it this way: There will never be a fractional number of ways to select groups of people. If the denominators of the fractions don’t all cancel out to 1, you may have either made an arithmetic mistake, or missed a chance to cancel out.