*The following is a guest post from Jonathan Taves, that was originally published on his website, EF Essays: Essays on Economics and Finance. *

With math on the GMAT, there’s always room for seconds. For those seeking a score higher than the 560 average, the following advanced concepts are important to learn. They’re less frequently tested than the concepts we discussed last week, so don’t spend hours and hours trying to master all iterations of combinatorics, for example. Competence, not mastery, should be the goal here.

However, these are tested, and so in that sense, aren’t really “advanced” at all. They’re essential. Unfortunately, these concepts are rarely covered in math review guides that aren’t GMAT-specific. Make sure to supplement your studying with lessons on Distance/Speed/Time, Probability, and Combinatorics (Permutations and Combinations).

**Distance/Speed/Time**

Of the advanced math concepts tested on the GMAT, Distance/Speed/Time (DST) problems would probably seem the most self-explanatory. Most people drive cars and need to estimate how long it will take to get from one location to another.

*“Let’s see…I just passed Staples, so there’s 18 miles left until Wadena. I’m driving about 70 mph, so I should be home in 15 minutes.”*

A simple example, to be sure, but that situation forms the foundation for DST problems. Your “distance” is 18 miles and your “speed” is 70 mph. The “time” it will take to travel that distance will be 15 minutes. If you prefer equations:

D = S * T |

18 = 70 * T |

Solve for T = 18/70 hours, or about 15 minutes. |

In general, the DST formula states that the distance you travel is the product of the speed you’re going and the length of time you’re in motion. Because this is an algebraic concept, once you have two of the three variables you can solve for the third – and find the correct answer. Solving the equation becomes more difficult when two subjects are introduced into the problem set.

It seems like a difficult task, but using the concept of “relative speed,” it isn’t at all. Relative speed is the mathematical term to describe using the DST formula to compare two different objects in motion. Its formula is the same, except that “S” now equals Relative Speed, which is the sum of each objects’ rate of speed. Let’s illustrate this point with an example:

*Two cyclists start at the same time from opposite ends of a course that’s forty-five miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long will it take for them to meet?*

Right off the bat you know that T will be the same for Cyclist A and Cyclist B. You also know what the total distance will be, forty-five miles, but not the individual distances for cyclist A and B. To solve we’ll manipulate the DST formula into D = (S * T) + (S * T).

D = (S * T) + (S * T) |

45 = (14T) + (16T) |

Solve for T = 1.5 hours, or 90 minutes. |

When the GMAT decides to turn its DST problems up a notch, it can be helpful to build a table to organize the variables you have and what you’re trying to solve for. Let’s illustrate this point with an example:

*A passenger train leaves the depot two hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the speed of the passenger train if it overtakes the freight train in three hours.*

Distance (Miles) | Avg. Speed (MPH) | Time (Hours) | |

Passenger Train | D | S | 3 |

Freight Train | D | S – 20 | 2 + 3 = 5 |

Total | ? | ? | ? |

Knowing that D must be equal for the passenger and freight trains and that the two trains are meeting at a certain point – and therefore should be set equal to each other – we can solve.

S * T = S * T |

S * 3 = (S – 20) * 5 |

Solve for S = 50 mph |

To conclude, there are two general points to remember when tackling more difficult DST problems, like the ones we worked above:

- First, when A & B are traveling in the same direction, the distance variable of the DST equation is the same.
- Second, when A & B are traveling in opposite directions, the time variable of the DST equation is the same.

If either of these cases apply, set the two equations equal to each other – this will allow you to quickly solve for the missing variable.

**Probability**

Probability is a number that expresses the likelihood of an event occurring. It can be solved by dividing the number of desired outcomes by the number of possible outcomes: P = n / N. In other words, the Probability of A plus the Probability of B must equal one. In short, there can never be a greater than 100% chance that two connected outcomes will occur. A basic example of this formula is as follows:

*Brian flips a coin once. What is the chance that it will turn up heads?*

P = n / N |

Solve for P = 1 / 2, or a probability of 50% |

Using the same concept, here’s a more difficult example:

*The probability is (1/2) that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?*

The easiest way to solve this problem is to do it in reverse. The probability of getting heads each time is (1/2)*(1/2)*(1/2) = (1/8). Therefore, using the property that “the Probability of A plus the Probability of B must equal one,” the probability that heads won’t show each time is 1 – (1/8). The answer is then (7/8).

A difficult example, yes, but more importantly, it brings up the concept of a probability being “independent vs. mutually exclusive.” Did you notice that we multiplied each probability together? Why did we do that and not add them, instead? When finding the probability of multiple events, individual events are added or multiplied depending on their relationship to each other.

Independent

An event is independent if its occurrence doesn’t influence other events. For example, if someone was tossing a coin and rolling a die, what is the probability he would get tails and roll a six? (1/2) * (1/6); a probability of (1/12).

Mutually Exclusive

On the other hand, if two events can’t occur at the same time, then they’re mutually exclusive. For example, is someone was rolling a die, what is the probability that he would get at least a three? You can’t get a three and something else at the same time – these events are mutually exclusive. (1/6) + (1/6) + (1/6) + (1/6); a probability of (2/3).

Another key concept of probability deals with replacement. In multiple event probability problems, whether or not the event can be “added back to the pool” and potentially recur is important to identify. For example, if you are trying to calculate the probability of picking a certain marble out of a bag, the first is (1/10), and if it isn’t replaced, then the second is (1/9), and so on. Of course if it is replaced, then both your first and second pick carry a probability of (1/10).

To illustrate, let’s work the following problem: *There are five married couples at a dinner party. If the chef was to only select three of the ten guests to test his food, what is the probability that none of them are married to each other?*

Step 1 – Independent or Mutually Exclusive

Independent because each dinner guest is being selected in three separate events.

Step 2 – Calculate Probabilities

Choice 1 | 10/10 (A person can’t be married to itself) |

Choice 2 | 8/9 (One couple and one person are eliminated) |

Choice 3 | 6/8 (Two couples and two people are eliminated) |

Step 3 – Combine Probabilities

P = (10/10) * (8/9) * (6/8)

P = (2/3)

**Combinatorics**

I’m sure you’ve heard of “saving the best for last.” In this case, we’ve saved the most difficult for last. Combinations and permutations, or the mathematical topic of combinatorics, cover the properties that characterize a set of numbers. The most basic derivative of combinatorics is enumeration. Enumeration is essentially a method of counting all possible ways to arrange elements.

As you know, the GMAT isn’t a test of mathematical skills, but a test of critical thinking. For that reason, sometimes the best approach to answering a problem is to use a basic form of mathematics. In this case: counting. For example, let’s say you’re asked how many ways one could arrange three marbles – one blue, one gray, and one green – in a row. If you physically draw out the possibilities, you’ll eventually come up with six:

1). |
BLUE | GRAY | GREEN | 4). |
GRAY | BLUE | GREEN |

2). |
BLUE | GREEN | GRAY | 5). |
GREEN | GRAY | BLUE |

3). |
GREEN | BLUE | GRAY | 6). |
GRAY | GREEN | BLUE |

For a basic problem, this is effective, but it’s right at the limit of what you can comfortably solve in the time limit for such a question on the GMAT – about two minutes. If you were more familiar with combinatorics, you’d realize this answer is equivalent to the factorial of 3. (Specifically, 3! = (1 * 2 * 3), or 6.) Knowledge of factorials is vital in combinatorics and is at the center of combination and permutation formulas.

- A combination is an
**unordered**collection of “k” objects taken out of “n” distinct objects: C(n, k) = n! / (k!(n! – k!)). - A permutation is an
**ordered**collection of “k” objects taken out of “n” distinct objects: P(n, k) = n! / (n! – k!).

When trying to set up a combination/permutation formula, think about the “k” variable as what you have and the “n” variable as what you want. The easiest way to show this is with a problem:

*The Granola Store’s owner, Sarah, wants to create a new website. If she has five graphic design companies to choose from and need to pick two of them to submit bids, how many different teams of design firms could she use?*

* A “combination” because the order in which the design firms are chosen is irrelevant. |

C(5, 2) = |

* This is the combination of what we have, five firms, to what we want, two firms. |

C(5, 2) = 5! / [2! * (5! – 2!)] = 10 |

* Sarah has ten potential sets of graphic design firms to choose from. |

For some of the larger combination/permutation problems you’re tasked with, a great way to simplify them is by canceling out factorials:

C(17, 14) = |

= 17! / 14! * (17! – 14!) |

* You can simplify the equation by cancelling out 14! from the numerator and denominator. |

= 17! / 3! = 680 |

The more difficult problems integrate more than one event into the question. These use similar concepts as probability problems in that “and vs. or” comes into play. As a reminder, “or” means to add; “and” means to multiply. Let’s illustrate with a couple of sample problems:

*A fraternity needs three seniors and two juniors to go on a beer raid. If the frat has six seniors and five juniors, how many different teams of frat bros can be sent?*

Step 1 – Ordered or Unordered

Unordered because it doesn’t matter what order they’re picked in, we just need to know if they’re going or not.

Step 2 – Set up Formulas

C(6, 3) = The combination of what we have, six seniors, to what we want, three seniors

C(5, 2) = The combination of what we have, five juniors, to what we want, two juniors

Step 3 – Combine Formulas

6! / 3!(6! – 3!) = 20

5! / 5!(5! – 2!) = 10

= Seniors and Juniors can be chosen go on the raid, so multiply the two products:

20 * 10 = 200 (different teams of frat bros to send on the raid)

*After arriving at the liquor store, the youngest frat bro must choose what kind of beer to buy for his fraternity brothers. The liquor store carries seven different kinds of beer. If he picks at most two, how many different combinations are possible?*

Step 1 – Ordered or Unordered

Unordered because it doesn’t matter what order they’re picked in, we just need to know what kinds of beers were or weren’t chosen.

Step 2 – Set up Formulas

C(7, 1) = The combination of what we have, seven kinds, and the frat bro picking at most two

C(7, 2) = The combination of what we have, seven kinds, and the frat bro picking at most two

Step 3 – Combine Formulas

7! / 1!(7! – 1!) = 7

7! / 2!(7! – 2!) = 21

= One or two kinds of beer can be purchased, so add the two products:

7 + 21 = 28 (different combinations if the frat bro picks at least two beer brands)

One last note on combinatorics: at their core they are counting problems. When approaching them on the GMAT, don’t overthink it. If you start trying to choose “Combination vs. Permutation” first, you might miss what the question is truly asking. The C & P formulas should only be used when necessary; it you can solve the problem logically or by enumeration, the chances of you making an accidental error will go down dramatically.