Counting Problems on the ACT and GRE Today I would like to start a thread on ‘counting.” Although counting sounds like it should be an easy concept, students tend to find math problems on standardized tests such as the ACT and GRE that involve counting among the most difficult. Writing A List Okay let’s start with the most basic strategy, namely ‘writing a list’… Sometimes the easiest way to count the number of possibilities is to simply list them all. When doing this it is important that you have a systematic way of forming your list. This will reduce the likelihood of missing something, or listing something twice. With that said, let’s look at a simple counting problem: Example 1 A menu lists 2 meals and 3 drinks. How many different meal drink combinations are possible from this menu? Some of you may already be able to solve this problem very quickly, but let’s first solve this by writing a list. Let’s make up 2 meals and 3 drinks, say chicken and fish for the meals, and water, soda, and juice for the drinks. Now let’s list all meal-drink combinations. To save time we will abbreviate each meal and drink by using its first letter. cw cs cj fw fs fj We see that there are 6 meal-drink combinations. Now, as many of you probably already realize, we can actually solve this problem very quickly by using the counting principle. (2)(3) = 6 The Counting Principle The counting principle says that if one event is followed by a second independent event, the number of possibilities is multiplied. The 2 events in the above problem are “choosing a meal,” and “choosing a drink.” Now let’s get a little more advanced and look at the counting principle in full generality. More generally, the counting principle says that if E1, E2, …, En are n independent events with m1, m2, …, mn possibilities, respectively, then event E1 followed by event E2, followed by event E3, …, followed by event En has m1∙m2 ∙∙∙ mn possibilities. My head is spinning after looking at all these E’s and m’s, so let’s try a problem to see this principle in action. Example 2 How many integers between 9 and 300 have the tens digit equal to 2, 3, or 4 and the units digit (ones digit) equal to 5 or 6? Let’s first solve this using the more general version of the counting principle: There are 2 possibilities for the ones digit (5 or 6). There are 3 possibilities for the tens digit (2, 3, or 4). There are 3 possibilities for the hundreds digit (0, 1, or 2). The counting principle says that we multiply the possibilities to get (2)(3)(3) = 18. That’s not so bad, is it? Note that we could also solve this problem by writing a list: Let’s simply list all the numbers in increasing order: 25 26 35 36 45 46 125 126 135 136 145 146 225 226 235 236 245 246 And that’s it. Once again we see that the answer is 18. Okay, let’s do a more difficult counting problem. This one is Level 5: Example 3 How many integers between 3000 and 4000 have digits that are all different and that increase from left to right? The best way to solve this is by writing a list: As you can see, there are 20 integers in this list. It would be good to take note that we only wrote down the necessary information when forming our list. For example, the second entry was just written “7” instead of “3457.” This will save a substantial amount of time. Also a clear and definite pattern was used in forming this list. In this case the list was written in increasing order. This will minimize the risk of duplicating or leaving out entries. Follow the advice given here with similar problems you see on the ACT or GRE and you will be able to answer them quickly and efficiently. For more advanced counting techniques take a look at my articles on Permutations and Combinations. Where To Find More Practice Problems If you are preparing for the ACT, GRE or any other standardized test that has counting problems, you may want to take a look at the Get 800 collection of test prep books. Click on the picture below for more information. And if you liked this article, please share it with your Facebook friends: Comments comments