Math for Liberal Arts

Solution 1: One way to solve this problem would be to systematically list each possible meal:

Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.

Solution 2: Another way to solve this problem would be to list all the possibilities in a table:

In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn't really care what the possible meals are, only how many possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It's always good when you solve a problem two different ways and get the same answer!)

Solution 3: We already have two perfectly good solutions. Why do we need a third? The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we've used the rows of the table for the appetizers and the columns for the main courses—where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn't terribly easy, we need a better way in case we have three categories to choose form instead of just two.

So, back to the problem in the example. What else can we do? Let's draw a tree diagram. We can abbreviate Breadsticks (B), Soup (SP), and Salad (SD) to make the tree easier to read.

This is called a "tree" diagram because at each stage we branch out, like the branches on a tree. In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer). We count the number of branches at the final level and get (surprise, surprise!) 15.

If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.

There are 21 choices from the first category and 18 for the second, so there are \(21\cdot 18=378\) possibilities.

There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are \(3 \cdot 5 \cdot 2=30\) possibilities.

There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in \(2\cdot 2\cdot 2=8\) different ways. Recall that another way to write \(2\cdot 2 \cdot 2\) is \(2^3\) which is much more compact.

1. There are 10 possible numbers (0 to 9) that could be in each of the four digits of the keycode.

   \begin{equation*} 10 \cdot 10 \cdot 10 \cdot 10 = 10^4 = 10,000 \end{equation*}

   There are 10,000 different keycodes.

2. There are 10 possible numbers (0 to 9) that could be in the first digit of the keycode. The second digit has to be different from the first, so that leave 9 possible numbers for the second digit. The third digit must be different from the first two digits, so there are 8 possible numbers for the thrid digit. Using the same reasoning, that leave 7 possibilities for the fourth digit, and:

   \begin{equation*} 10 \cdot 9 \cdot 8 \cdot 7 = 5,040 \end{equation*}

   There are 5,040 keycodes that have no repeated digits.

3. Since all of the digits have to be odd numbers (1, 3, 5, 7, 9), there are 5 possible numbers that could be in each of the four digits of the keycode.

   \begin{equation*} 5 \cdot 5 \cdot 5 \cdot 5 = 5^4 = 625 \end{equation*}

   There are 625 keycodes that only contain odd numbers.