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I was recently discussing a particular GMAT problem with a friend, and as so often happens with standardized-test nerds, the discussion turned into an extended analysis. We can’t help ourselves, I suppose.

The question went something like this:

*Jim and John are workers in a department that has a total of six employees. Their boss decides that two workers from the department will be picked at random to participate in a company interview. What is the probability that both Jim and John are chosen?*

Now, with many GMAT problems, there are multiple ways to skin a cat. As it turns out, my friend and I chose completely different strategies that arrived at the same answer. But interestingly enough, our different strategies got us to hit upon some key distinctions between **probability **and **combinatorics**.

**1.** My friend chose to go strictly with probability:

There is a 1/6 chance that Jim will be selected first. Then, there are 5 workers left, so the probability that John is chosen next is 1/5. Therefore, the probability of Jim being chosen first, then John being chosen second is simply 1/6 * 1/5 = 1/30.

However, we also have to consider the possibility that John is chosen first and Jim second. That still leads to the same number: 1/6 * 1/5 = 1/30.

So, because we are interested in each of these possibilities (and nothing else), we must add the two probabilities to get the final answer:

1/30 + 1/30 = **1/15**.

**2.** I chose to bring combinatorics into the picture:

There are 15 possible combinations of 2 people that you can choose from a group of 6. You can find this using the combination formula:

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

In this case, n = 6, since there are six people total, and k = 2, since we’re finding a subgroup of two. Therefore:

6! / (2! * 4!) = 6 * 5 / 2 = 15 total combinations.

Now, out of those 15 combinations, we are interested in only one — Jim and John. And recall that this is a combination (where order does not matter), as opposed to a permutation (where order does matter). Jim and John is the same combination as John and Jim, since the same two people are involved.

(For clarification, it would be a permutation if, say, John and Jim were running a race, and we awarded different prizes for 1st and 2nd. In that case, Jim finishing first is different from John finishing first. But in our problem, we’re not concerned with who is picked first; we only care about who’s in the group of two).

Back to the problem: We’re interested in only one combination (Jim and John) out of a total of 15. Therefore, the final answer is **1/15**.

“But wait,” said my friend, “It’s a combination, so that means order shouldn’t matter. Jim and John is the same combination as John and Jim. So how come in my solution, we added two different probabilities for Jim-John and John-Jim? Order shouldn’t matter, but in my solution, it did.”

What we realized is that order mattered in my friend’s solution because he was considering two different **events**, not two different **combinations**. Jim and John is the same combination as John and Jim, so if we were restricting ourselves to finding information solely about combinations, then order would not matter.

However, we were not only finding information about combinations; we were also interested in **probability**. The situation of Jim and John being chosen first and second, respectively, is a **distinct event** from that of John and Jim being chosen first and second, respectively. So even though both events involve the same **combination **of people, the **events **themselves are different.

What makes problems like this a little bit tricky is that they can involve **both **probability and combinatorics, and it might be easy to confuse the two. But always remember, combinatorics on their own deal solely with finding *the number of combinations or permutations* in a given set of data, while probability deals with finding *the likelihood that an event or events will occur*.

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