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Combinations and . . . Non-combinations – Part II

As I stated in my last post, the Fundamental Counting Principle (FCP) can be used to solve the majority of counting questions on the GRE.

The FCP says:

If we have a task consisting of stages, where one stage can be accomplished in A ways, another stage in B ways another in C ways . . . etc., then the total number of ways to accomplish the entire task will equal A×B×C×… 

The great thing about the FCP is that it’s easy to use, and it doesn’t require the memorization of any formulas. So, whenever I encounter a counting question, I first try to determine whether or not the question can be solved using the FCP. To determine this, I ask, “Can I take the required task and break it into individual stages?” If the answer is yes, I may be able to use the FCP to solve the question.

To see how this plays out, let’s solve the following question:

How many different 3-digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?

(A) 222

(B) 245

(C) 291

(D) 315

(E) 343

So, our task is to find 3-digit numbers that adhere to some specific rules. Can we take this task and break it into individual stages? Sure, we can define the stages as:

Stage 1: Choose a digit for the hundreds position

Stage 2: Choose a digit for the tens position

Stage 3: Choose a digit for the units position

Once we accomplish all 3 stages, we will have “built” our 3-digit number.

At this point, we need to determine the number of ways to accomplish each stage.

Stage 1: In how many different ways can we choose a digit for the hundreds position? Well, since the 3-digit number must be greater than 299, the digit in the hundreds position cannot be 0, 1 or 2. The question also says that the digits 6 and 8 are forbidden. So, when we consider the various restrictions, we see that the digit in the hundreds position can be 3, 4, 5, 7 or 9. So, there are 5 different ways in which we can accomplish Stage 1.

Stage 2: In how many different ways can we choose a digit for the tens position? Well, since the tens digit can be any digit other than 1, 6 or 8, we can see that the tens digit can be 0, 2, 3, 4, 5, 7 or 9. So, there are 7 different ways in which we can accomplish Stage 2.

Stage 3: The units digit can be 0, 2, 3, 4, 5, 7 or 9. So, there are 7 different ways in which we can accomplish Stage 3.

At this point we can apply the FCP to see that the total number of ways to accomplish all three stages (and create our 3-digit numbers) will equal the product of the number of ways to accomplish each individual stage.

So, we get 5 × 7 × 7, which equals 245.

There are 245 different 3-digit numbers that are greater than 299 and do not contain any 1’s, 6’s or 8’s.  The answer to the original question is B.

In the next post, we’ll examine some additional considerations when applying the FCP.

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