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Knewton Tips: Probability vs. Combinatorics on the GMAT

*Rich*

*Zwelling*

*is one of*

*Knewton's*

*expert*

*GMAT prep*

*instructors. He enjoys using the word “*

*com*

**

*bi*

**

*na*

**

*torics*

*” in everyday conversation.*

–

A friend and I were recently analyzing a particular GMAT problem. Since we’re both standardized-test geeks, the discussion naturally turned into an extended investigation into GMAT strategies. Some topics are just too fascinating to avoid!

The problem in question went like this:

*Melinda and Mindy both work in a department consisting of six total employees. Their boss specifies that two workers in the department will be chosen at random to take part in a company survey. What’s the probability that both Melinda and Mindy will be chosen for the survey?*

As with many GMAT problems, this problem can be solved in a variety of ways. Turns out that my friend and I chose entirely different methods—but we both ultimately arrived at the same (correct!) answer. As it turns out, our divergent strategies highlighted some key distinctions between

**probability**and**combinatorics**as used on the GMAT.1. My friend decided to go with

**probability**to solve the problem. Here’s his thought process:The chance that Melinda will be selected first is 1/6. If that happens, there will be 5 workers left. The probability that Mindy is the next person chosen is 1/5. Thus, the probability of Melinda being the first person chosen, and Mindy being the second person chosen, can be found by multiply 1/6 * 1/5 = 1/30.

In addition, however, we must factor in the possibility that Mindy is the first person chosen and Melinda the second. This will lead to the same probability: 1/6 * 1/5 = 1/30.

Since we’re only interested in these two possibilities (and nothing else), we add the two probabilities in order to arrive at our final answer.

1/30 + 1/30 =

**1/15**. This is the probability that both Melinda and Mindy will be chosen for the company-wide survey.2. Unlike my friend, I decided to use

**combinatorics**to solve the problem. Here’s my train of thought:With a group of 6 people, there are 15 possible combinations of 2 people that you can choose. This can be calculated using the combination formula, namely:

n! / [k! * (n-k)!]

In this problem, n = 6, because there are six people in all, and k = 2, since we’re looking for a subgroup of two people. Therefore, we can figure out that:

6! / (2! * 4!) = 6 * 5 / 2 =

**15**total combinations of 2 people.So, out of these 15 possible combinations, we’re only interested in one: Melinda and Mindy. Remember, since this is a combination, order does

**not**matter (as opposed to a permutation, where order must be taken into account). Melinda and Mindy is the exact same combination as Mindy and Melinda, since the same two people are involved.(To explain this further: an example of a permutation would be if Melinda and Mindy were in a cycling race, and there were different prizes awarded for 1

^{st}and 2^{nd}places. In that case, Melinda finishing first is a different scenario from Mindy finishing first. But in our problem, we don’t care who’s picked first, but only about who is in the group of 2; therefore, we don’t need to worry about order.)So, back to the question. We’re interested in only one combination, Melinda and Mindy, out of a total of 15 combinations. Therefore, the final answer is

**1/15**—the same answer that my friend came up with using probability.When we talked about this, though, my friend interrupted me. “But wait,” he said, “since it’s a combination, order shouldn’t matter, right? Melinda and Mindy is the same exact combination as Mindy and Melinda. So—how come in my solution, we added different probabilities for Melinda-Mindy and Mindy-Melinda? Order shouldn’t matter here—but it did!”

After a little bit of discussion, we realized that the order had mattered in my friend’s solution because he had looked at the situation as two different

**events**, not two different**combinations.**Melinda and Mindy**is**the same combination as Mindy and Melinda—so if we were restricting ourselves to finding information that was solely about combinations, then order wouldn’t matter.However, in this case, we were also interested in

**probability.**The situation of Mindy and Melissa being chosen first and second, respectively, for the survey, is a**distinct event**from if Melissa and Mindy were chosen first and second. So, even though we know that both events concern the same**combination**of people, the events are different.Problems like this can be a little bit hard to follow, as they can involve

**both**probability and combinatorics, making it easy to confuse the two. It’s important to remember that, on their own,**combinatorics**deal only with finding the*number of combinations or permutations*in a given set of data, while**probability**deals with discerning*the likelihood that an event or events will happen.*