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GMAT Combinatorics 1.2: Problem Solutions

Posted by Mark Skoskiewicz on Tue, Dec 27, 2011 @ 11:03 AM

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1. Suppose you perform the following experiment - You flip a coin 5 times and record the results. How many different outcomes are there? (A result being a 5-character long string of H's and T's, such as HHTHT)

Solution: For every flip of the coin, there are two possible outcomes. Because we flip the coin 5 times, according to the fundamental principle of counting the number of possible outcomes is:

[latex]2 \times 2 \times 2 \times 2 \times 2 = 32[/latex]

Note that this calculation can also be expressed by [latex] 2^5 [/latex]. In fact, if you flip a coin [latex] n [/latex] times, the number of possible outcomes is [latex] 2^n [/latex].


2.0 You receive a new debit card in the mail and you are asked to create a new PIN (Personal Identification Number) for your debit card. The PIN must be 4 digits long and each digit must be one of the numbers 0 - 9. You can use the same digit more than once. How many possible PIN numbers are there?

Solution: Each number in your PIN can be one of ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. One digit must be assigned on each of four places. So, by the fundamental counting principle, the number of outcomes is:

[latex]10 \times 10 \times 10 \times 10 = 10,000 [/latex]


2.1 Same as 2, but now you can use each digit only once.

Solution: Now that we cannot reuse digits, once we assign a digit to the first place, we will only have 9 options for the second place, then 8 options for the 3rd place and so on. As I mentioned before, this is a permutation - the ordering or a subset of elements taken from a larger set. But, as you'll see in the course of these posts on combinatorics, all the technical jargon and formulas you get in other prep material just reduces to a single principle - the fundamental principle of counting. The number of possible PIN numbers is now,

[latex] 10 \times 9 \times 8 \times 7 \times [/latex] = 5040


2.2 Same as 2, but now you can use each digit only once and you cannot begin the PIN with a zero.

Solution: This is very similar to the last problem with one exception. We only have nine possibilities for the first digit. So the number of possible PIN numbers is

[latex]9 \times 9 \times 8 \times 7 = 4536 [/latex]


3. Sam, Sally, Xanthar, and their parents drive to the park in their Prion (a Prius that has been modified by Xanthar for hydrogen ion technology). If the front of the car has two seats (the drivers seat and the passenger seat), the back seat will hold three people, and the two parents sit in the front and Sam, Sally, and Xanthar sit in the back seat, how many ways can the family be arranged in the car if each seat contains only one passenger.

Solution: The total arrangements in the car depends on the arrangements of two groups - the kids and the parents. We will take the product of these two arrangements because they are not mutually exclusive.

[latex]Kid Arrangements = 3 \times 2 \times 1 = 6[/latex]

[latex]Parent Arrangements = 2 \times 1 = 2[/latex]

[latex] Total Arrangements = 6 \times 2 = 12[/latex]


4.0 You roll two die. How many possible outcomes are there?

Solution: You can think of the outcome of one roll as an ordered pair or numbers [latex](x, y)[/latex]. For instance, the ways to roll a sum of seven are [latex](1, 6), (2, 5), (3, 4), (4, 3), (5, 2),[/latex] and [latex](6, 1)[/latex]. Obviously, there are six options for the first number in the ordered pair, and six options for the second number in the ordered pair. So, according to the fundamental principle of counting the number of outcomes is:

[latex]6 \times 6 = 36[/latex]


4.1 Same as 4.0. How many ways can you get two even numbers?

Solution: We can use the same thinking as before. However, this time, because we are only considering even numbers, there are only 3 options for each number. Thus, the number of possibilities is

[latex]3 \times 3 = 9[/latex]


4.2 Same as 4.0. How many ways can you get two odd numbers?

Solution: This is the exact same problem as the one above. See the solution to 4.1.


4.3 Same as 4.0. How many ways can you get one even and one odd number?

Solution: This may seem somewhat complicated as we don't know in what order the even number and odd number appear in our ordered pair. However, if you were paying attention when you read my original combinatorics post, you might be thinking about the 'complement method.' In fact, it's perfect for this problem. On any given roll, one of three things will happen:

    • You'll get two odds

    • You'll get two evens

    • You'll get an even and an odd

We already know the number of ways in which the first two outcomes can happen, and we know the total number of outcomes. So, we can use the complement method:

Even & Odd = Total - Two Odds - Two Evens

Even & Odd = 36 - 9 - 9 = 18


4.4 Same as 4.0. Now take the sum of the two upward facing numbers. How many ways can the sum be an even number?

Solution: We can only get and even sum in two ways:

    • Even + Even = Even

    • Odd + Odd = Even

Using the problems we already solved above, we know that there are nine ways to roll two evens and nine ways to roll two odds. Since these are mutually exclusive events we add to get the total number of outcomes, which is 18.


4.5 Same as 4.0. Now take the product of the two upward facing numbers. How many ways can the product be an even number?

Solution: We proceed as above. We get an even product in one of two ways

    • Even X Even = Even

    • Even X Odd = Odd X Even = Even

The number of ways the first even can happen is nine and the number of ways the second event can happen is 18. Since they are mutually exclusive, we add, and get 27.

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