How many three-digit positive integers are there such that the

How many three-digit positive integers are there such that the three digits are $3$, $4$, and $8$ ?

  1. $3$
  2. $6$
  3. $9$
  4. $15$
  5. $96$

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 8 of Section 6 of Practice Test 1. Those questions testing our knowledge of Counting Methods can be kind of tricky, but never fear, PrepScholar has got your back!

Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.

The question asks us to count the number of three-digit integers meeting certain criteria, so it likely tests our Counting Methods math skill. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.

What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

 

  1. We want to know how many three-digit positive integers there are with the digits $3$, $4$, and $8$

 

Develop a Plan

We want to form three-digit integers using three distinct digits. So we have three blanks to fill with our integers:

$$\;\;\;?\;\;\;\;\;\;?\;\;\;\;\;\;?$$

Thinking of this logically, we have three different options for the hundreds integer, any of the three integers $3$, $4$, or $8$:

$$\;\;\;\3\Possibilities\;\;\;\;\;\;?\;\;\;\;\;\;?$$

For the next integer, we should realize that we’ve already used one of our integers, so we only have two options remaining:

$$\;\;\;\3\Possibilities\;\;\;\;\;\;2\Possibilities\;\;\;\;\;\;?$$

Finally, we’ll have just one integer remaining for the last spot after filling in the integers in the hundreds and tens places:

$$\;\;\;\3\Possibilities\;\;\;\;\;\;2\Possibilities\;\;\;\;\;\;1\Possibility$$

Now let’s use our knowledge of Counting Methods to finish solving this question.

Solve the Question

To determine the number of possibilities, or permutations, for different events that occur sequentially, we multiply together the number of possibilities for each of those events. In other words, for counting the number of permutations in this question, we should multiply $3$ possibilities by $2$ possibilities by $1$ possibility:

$$(3\Possibilities)·(2\Possibilities)·(1\Possibility)=3·2·1 = 6$$

So there are Six three-digit positive integers that can be formed from the digits $3$, $4$, and $8$. In fact, we could list them all if we really wanted to: $348, 384, 438, 483, 834, \and 843$. The correct answer is B, $6$.

What Did We Learn

For counting questions where the order of the individual items matters, we can think of the solution as the product of all of the possibilities for each item. Additionally, it was important to realize that we could not repeat any digits, so after using one digit in the hundreds place, then we only have two digits remaining for the tens place.

 

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