Combinatorics

2.1.7.1 There are 4 choices for colour; 2 choices for air conditioning; 5 choices for stereo; and 3 choices for floor mats. So in total, there are \(\displaystyle 4 \cdot 2 \cdot 5 \cdot 3=120\) different combinations of options. Of these, \(\displaystyle 3\) combinations are available at the dealership, so the probability that one of these cars has the options Ace wants is \(\displaystyle 3/120=1/40\text{.}\)

2.1.7.2 In Candyce's book, the reader will have \(\displaystyle 3\) choices at the first decision point, and \(\displaystyle 2\) choices at each of the following three decision points. Thus, there are a total of \(\displaystyle 3\cdot 2\cdot 2 \cdot 2=3\cdot 2^3=24\) possible storylines. Candyce must write \(\displaystyle 24\) endings.

2.2.6.1 Let's call the black markers Black A, Black B, and Black C. The four possible options are: I leave behind the blue marker; I leave behind Black A; I leave behind Black B; I leave behind Black C. In each of the final three cases, I take the blue marker. Therefore, the probability that I take the blue marker is \(\displaystyle (1+1+1)/4=3/4\text{.}\)

2.2.6.2 If Ocean is thinking of a letter, there are \(\displaystyle 26\) things they could be thinking of. If they are thinking of a digit, there are \(\displaystyle 10\) things they could be thinking of. In total, there are \(\displaystyle 10+26=36\) things they could be thinking of.

2.3.4.1 We divide the possible passwords into three cases, depending on whether the digit is in the first, second, or third position. In each case, we have \(\displaystyle 10\) choices for the digit, \(\displaystyle 26\) choices for the first lowercase letter, and \(\displaystyle 26\) choices for the second lowercase letter. Thus, in each case we have \(\displaystyle 10\cdot 26^2\) possible passwords. In total, there are \(\displaystyle 10\cdot 26^2+10\cdot 26^2+10\cdot26^2=30\cdot 26^2=20,280\) different passwords with these constraints.

2.3.4.2 We divide the possible passwords into two cases, depending on whether there are 8 or 9 characters. If there are \(\displaystyle 8\) characters, then the product rule tells us that there are \(\displaystyle 10^8\) passwords consisting entirely of digits (\(\displaystyle 10\) choices for the digit in each of the \(\displaystyle 8\) positions). Similarly, if there are \(\displaystyle 9\) characters, then there are \(\displaystyle 10^9\) passwords consisting entirely of digits. In total, there are \(\displaystyle 10^8+10^9=1,100,000,000\) passwords with these constraints.

2.4.1.1 Use both rules. There are two cases that should be added together: the number of numbers that have two digits, and the number of numbers that have four digits. For each of these cases, use the product rule to determine how many numbers have this property. (The answer is \(\displaystyle 9 \cdot 10+9\cdot 10^3=9,090\text{.}\) Note that in order for a number to have exactly \(\displaystyle k\) digits, the leading digit cannot be zero.)

2.4.1.2 Use only the product rule. There are \(\displaystyle 6\) outcomes from the red die, and for each of these, there are \(\displaystyle 6\) outcomes from the yellow die, for a total of \(\displaystyle 6\cdot 6=36\) outcomes.