## How to Master the Trickiest Series of Questions on the GRE

Posted on March 2, 2015 | Filed in GRE

Find out how to solve probability, combination, and permutation problems on the GRE exam by paying attention to keywords. Probability, combination, and permutation problems on the GRE pose a lot of stress on students, but learn how to identify the best methods to solving each question with these key tips.

Math equations can be solved by utilizing appropriate formulas, following easy to memorize rules. The verbal section requires concentration and a mastery of English standard grammar. There exists a series of questions on the GRE, however, that require the use of all of your test-taking senses. Probability, combination, and permutation questions require an instinct for flags, flags that give away tell tale signs as to how to solve for the solution. Once you understand exactly what the question is asking for and why, the problem will seem as flipping a coin.

1. Mutually Exclusive or Independent

It is important to remember the difference between factors that are mutually exclusive and factors that are independent of each other. Say you are given ten boxes of chocolates, each box containing a different flavor. Imagine that chocolates containing nuts can only be found in boxes 6 through 10. If you would be asked what is the probability of selecting a chocolate from one of the first 5 boxes AND have the chocolate contain nuts, the resulting answer would have to be zero because boxes 1 through 5 don’ contain nuts. The two factors are mutually exclusive. If you were asked to flip a coin followed by rolling a 6 sides dye. The probability of landing on heads and rolling on a 6 are possible because the outcome of one event does not affect the outcome of the other. The two events are independent of each other.

2. The Difference Between AND and OR

There is a huge difference between the probability of two events occurring, or either of two events occurring. If you are asked to roll a die and asked what is the probability of rolling an even number and a number less than 4, you are still being asked the probability of a single event. This single event just happens to have more than one qualifier. There are 3 even numbers on a die, and 3 numbers less than 4. The number 2 is the only digit on the die that meets both of the given requirements. There is one outcome, out of the six total possible outcomes, that would work for our needs. Now say you were asked what the probability of rolling a number that is either a 4 or an odd digit. We are now asked to find the probability of two separate events. We are asked to find the probability of rolling a 4, and we are also asked what the probability is of rolling an odd digit. There is one outcome resulting in a rolled 4 and 3 possible outcomes for rolling an odd digit. Four outcomes would meet our requirements out of a total six possible outcomes.

3. Permutation and Combination

Always be asking yourself, how important is order? When entering the code into a pad lock the order of the digits is very important. If the right code is 144, 414 would not open the lock. This kind of problem is referred to as a permutation problem. There are 10 possible choices for each digit on the lock (in the case of this three digit lock we would need to multiple 10 by 10 by 10 to see the total number of permutations). You can also have a permutation problem in which repetition is not allowed. Say that once a number is used on that imaginary lock we discussed, it can no longer be used. In that case the options change (the first digit has 10 options, while the second has 9 as a result of one outcome being removed as a possibility, the next digit has 8 possible outcomes). To solve this sort of permutation problem we would multiple 10 by 9 by 8. A combination problem is one in which the order of the factors doesn’ matter. A lottery is a combination problem without repetition. The order of the numbers doesn’ matter and numbers can appear more than once. Say this particular lottery we’re playing contains 6 numbers ranging from 01 up to 49. There are 49 possible outcomes for each number drawn. Because order doesn’ matter we need to reduce the amount of outcomes that would exist in a permutation problem. The formula is 49!/6!(49-6)!

We have 49 options for each number and only six numbers are going to be drawn. Combination problems with repetition are a bit trickier. Imagine you are reaching into a refrigerator filled with 5 different types of sodas. If you would like to select only 4 sodas to take out, how many combinations of different sodas can be selected. The solution is derived from the following formula: (5+4 -1)!/4!(5-1)!

The reasoning is that all four sodas could be the same, or all different, or two the same, or three, and depending on which sodas are selected or excluded altogether.

The probability, permutation, and combination questions pose a lot of stress on test takers because they are very large in scope and design. They are questions about options and too many options can confuse anyone. The key is to visualize the problem, see it as a relatable situation, whether a combination lock, a lottery, or a refrigerator filled with sodas.