Does Order Matter? Combinations and Non-Combinations – Part III

In a recent article, I discussed my concern regarding the term “permutation.” In this article, I’ll discuss my concern regarding the question, “Does order matter?”  People often use this question to help determine whether to use permutations or combinations. The premise is that we use permutations when order matters, and we use combinations when order does not matter.

 

Unfortunately, the “Does order matter” question is not without its problems. First, as I already noted in that previous article, true permutation questions are rare on the GMAT. Second, the question “Does order matter?” can have unintended results. Consider this example.

 

There are 7 students. In how many ways can we elect a President and Treasurer, if no student can hold both positions?

 

Does the order of the selected students matter?

 

Some students might answer “No” to this question. Consider these election results:

 

Results #1: Ann is elected President, and Bob is elected Treasurer.

Results #2: Bob is elected Treasurer, and Ann is elected President.

 

Since these results are identical, we can safely conclude that order does not matter, which means that we can use combinations, right?

 

No.

 

The problem is that it’s not 100% clear what we’re asking when we ask, “Does order matter?” The issue becomes even more muddled when we tackle difficult, multi-step questions. Fortunately, the Fundamental Counting Principle (FCP) can help.

 

Aside: I’m going to assume that you’re familiar with the FCP. If you’re not familiar with it, please read this Beat The GMAT article.

 

When we apply the FCP, we first take a certain task and break it into stages. At that point, we should ask, “Does the outcome of each stage differ from the outcomes of the other stages?”

 

If the answer to this question is YES, then we can continue solving the question using the FCP. If the answer to this question is NO, we cannot solve the question using the FCP and, in most cases, we can use combinations.

 

Okay, now onto the question at hand.

 

There are 7 students. In how many ways can we elect a President and Treasurer, if no student can hold both positions?

We’ll take the task of “building” election results, and we’ll break it into individual stages:

Stage 1: Choose a President

Stage 2: Choose a Treasurer

Now ask, Does the outcome of each stage differ from the outcomes of the other stages?”

 

The answer is YES. The outcome of stage 1 (electing a President) is different from the outcome of stage 2 (electing a Treasurer). So, we’ll continue applying the FCP.

 

We can complete stage 1 in 7 ways, and we can complete stage 2 in 6 ways. By the FCP, we can complete both stages (and thus elect a President and Treasurer) in 7 x 6 ways (42 ways).   

 

Now let’s examine a different question.

 

Kyle has 7 students. In how many ways can he select 2 people to attend his party?

We’ll take the task of “building” a party guest list, and we’ll break it into individual stages:

Stage 1: Choose one person to attend the party

Stage 2: Choose another person to attend the party

Now ask, Does the outcome of each stage differ from the outcomes of the other stages?”

 

The answer is NO. The outcome of stage 1 (being a party guest) is the same as the outcome of stage 2 (being a party guest). So, we cannot apply the FCP to this question. Instead, we can use combinations. We can select 2 people from 7 people in 7C2 ways (21 ways)

 

Aside: If anyone is interested, we have a free video on calculating combinations (like 7C2) in your head.

 

Now let’s try a harder question.

 

From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an odd integer?

 

Let’s first see what happens when we ask, “Does the order of the two selected numbers matter?”

 

Well, selecting 7 then 20 (for an odd sum of 27) is the same as selecting 20 then selecting 7. Since the order doesn’t matter, this must be a combination question, right?

 

No.

 

Now let’s see what happens when we apply the FCP.

 

We’ll take the task of selecting 2 numbers (with an odd sum), and we’ll break it into stages. One way to do this is to first recognize that, for the sum to be odd, one number must be even, and the other number must be odd. So, our two stages could look like this:

Stage 1: Select an even number from the set

Stage 2: Select an odd number from the set

Does the outcome of one stage differ from the outcome of the other stage?

 

YES. In one stage we’re selecting an even number, and in the other stage, we’re selecting an odd number. Since the outcomes are different, we can continue applying the FCP.

 

We can complete stage 1 in 50 ways (since there are 50 even numbers in the set), and we can complete stage 2 in 50 ways (since there are 50 odd numbers in the set). By the FCP, we can complete both stages (and thus select 2 numbers with an odd sum) in 50 x 50 ways (2500 ways).  

 

Now let’s change things slightly.

 

From the set of integers from 1 to 100 inclusive, in how many ways can we select two different numbers such that their sum is an EVEN integer?

 

To solve this question, we must first recognize that, for the sum to be even, both selected numbers must be even OR both selected numbers must be odd. So, we have two different cases to consider.

 

Case a: both selected numbers are even: We’ll take the task of selecting two even numbers and break it into stages.

Stage 1: Select an even number from the set

Stage 2: Select another even number from the set

Does the outcome of one stage differ from the outcome of the other stage?

 

NO. In both stages, the outcome is the same. In both cases, we’re selecting an even number from the 50 even numbers in the set. So, for example, selecting 16 in stage 1 and selecting 54 in stage 2 is THE SAME as selecting 54 in stage 1 and selecting 16 in stage 2.

 

Since the outcomes of the two stages are the same, we cannot apply the FCP. Instead, we can use combinations. We can select 2 different even numbers from 50 even numbers in 50C2 ways (1225 ways). Now we’ll deal with the other possible case,

 

Case b: both selected numbers are odd: The steps here will be the same as those above. So, we can select 2 different odd numbers from 50 odd numbers in 50C2 ways (1225 ways).

 

So, the total number of ways we can select 2 numbers with an even sum equals 1225 + 1225 (= 2450).

 

Important Takeaways

 

When answering counting questions, you should first determine whether the required task can be broken into individual stages. If it is possible, then do so.

 

Once the task has been broken into stages, ask,“Do the outcomes of each stage differ from the outcomes of other stages?”

 

If the answer is YES, determine the number of ways to complete each stage, and then apply the FCP. If the answer is NO, then you most likely have a combination question.

 

There are additional strategies to consider, but the above steps will help you answer most counting questions on the GMAT.

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