All GMAT Math Resources
Example Questions
Example Question #21 : Counting Methods
In how many ways can I distribute four different presents to four different people, assuming each person only receives one present?
This problem is asking us in how many ways the four different presents can be "arranged," or in this case, given to different people. Therefore, the answer is :
Example Question #22 : Counting Methods
In how many ways can the letters of the word "Tennessee" be arranged to form a nine-letter combination?
The first thing that we should do is break down the different group of same letters in Tennessee: We have one "T," four "E's," two "N's," and two "S's." We have a total of nine letters; four of these are the same, as are two more pairs. In any combination problem, if we have a total of letters, then for every number of the same letters, we have . So, we can model this problem's situation with the expression :
Example Question #23 : Counting Methods
In how many ways can three of the same type of bike and two different skateboards be given to five people if each person only receives one item?
This question asks us to count the number of ways to arrange 3 of the same items and 2 different items. Let's assign letters to each of them: "" for one of the bicycles, for the first type of skateboard, and for the second type of skateboard. In this notation, our set looks like this:
We have a total of five individuals to receive five items, one item each. In any combination problem, if we have a total of letters, then for every number of the same letters, we have . So, for this situation, the total is given by dividing the total number of ways to arrange the items 5! by 3!, since 3 of these items are the same:
Example Question #22 : Counting Methods
In how many ways can the letters , , , and be arranged to form a three-letter combination if order is not important?
Here, the number of "slots" is different than the number of letters we can use. We must first determine in how many ways we can select the letters to be arranged in our three-letter combination. Because we are selecting some but not all out of a presented group of options and order is not important, we can use the formula , where is the total number of things from which we are choosing and is the number things we are selecting for each group. In this case, and , so our equation will look like this:
Example Question #24 : Counting Methods
We want to create a two-character code to reference items in our warehouse. We can use any single-digit number and any letter in the English alphabet to do so. How many different codes can be created?
We have 36 characters to choose from, since there are 10 single-digit numbers (0-9) and 26 letters in the alphabet. The length of the code is two characters and nothing was said about repeated combinations not being allowed, so the correct answer is given by , which is .
Example Question #25 : Counting Methods
In how many ways can the letters , , , , and be arranged to form a four-letter combination in which order is significant?
In this problem, we are selecting four letters from a group of five letters that contains a pair of duplicate letters: two s. We can model this situation using the formula , where is the number of things we have to choose from and each value of corresponds to the number of duplicate items in a given group. In this case, we have five letters to pick from, so , and two of them are duplicates, so . This makes our equation:
Example Question #26 : Counting Methods
We want to create a five-character-long password using only numerical digits through . How many different passwords can we create?
For each of the five "slots" that we want to have for our password, we have ten different digits to choose from. The number of passwords is given by calculating or , which equals .
Example Question #22 : Counting Methods
We want to create a three-character-long password using only the twenty-six letters from the English alphabet. How many different passwords can be created?
A three-letter-long password can be created in ways, since there are 26 letters to choose from and the password is three letters long.
Example Question #23 : Understanding Counting Methods
We have four keys and are asked to open four locks with those keys. Each key only opens one lock; however, we don't know which key opens which lock. At most, how many attempts must we make before we can be sure to have all locks opened?
We are asked to find the number of ways in which the keys can be used on the locks. There are four keys and four locks. This is just like looking for the number of permutations of four different letters. Therefore, it is given by 4!
Indeed, if we start with the first lock, we have four keys, and if we only use the right key last, then we have made four attempts on the first lock. Assuming that we leave the key in the first lock, we then have three keys remaining for the second lock. Similarly, if we pick the wrong key twice, then the third opens the lock, and so on and so forth. We can then multiply all these possibilities to obtain .
Example Question #30 : Counting Methods
A college cafeteria offers three sizes of pizza - small, medium, or large. With a small pizza, up to one topping is included without additional charge; for a medium or large, up to two different toppings are included without additional charge.
The cafeteria offers two meat toppings - pepperoni and sausage. If the cafeteria offers other toppings, then how many ways can someone order a pizza - choosing a size and up to the maximum number of toppings - without having to pay extra?
Assume that double toppings are not an option.
If a small pizza is chosen, one topping out of can be ordered for no additional charge.
If a medium pizza is ordered, up to two different toppings out of can be ordered for no additional charge - this is
If a large pizza is ordered, up to two different toppings can be ordered for no additional charge - again, this numbers .
Add these expressions: