Counting Principle
If two jobs need to be completed and there are m ways to do the first job and n ways to do the second job, then there are m × n ways to do one job followed by the other. This principle can be extended to any number of jobs.
EXAMPLE 7: How many integers between 100 and 1,000 consist of three different odd digits?
You are looking for 3-digit numbers, such as 135, 571, and 975, all of whose digits are odd and all of which are different. The best way to answer this question is to use the counting principle. Think of writing a 3-digit number as three jobs that need to be done. The first job is to select one of the five odd digits (1, 3, 5, 7, 9) and use it as the digit in the hundreds place. This can be done in 5 ways. The second job is to select one of the four remaining odd digits for the tens place. This can be done in 4 ways. Finally, the third job is to select one of the three odd digits not yet used to be the digit in the ones place. This can be done in 3 ways. By using the counting principle, the total number of ways to write a 3-digit number with three different odd digits is 5 × 4 × 3 = 60.
EXAMPLE 8: At the corporate headquarters of a large company, outside each office there is a keypad like the one in the figure below. To gain entrance to an office, an employee must punch in an access code and then press the enter button. If a code consists of the letter A or B followed by either two or three different digits, how many different access codes can be issued?
Again, we will use the counting principle. We have to calculate the number of two-digit codes and three-digit codes separately and then add them. For a two-digit code, there are 2 choices for the letter (A or B), 10 choices for the first digit (0 through 9), and 9 choices for the second digit (0 through 9, except for the digit already chosen). So, there are 2 × 10 × 9 = 180 two-digit codes. Similarly, there are 2 × 10 × 9 × 8 = 1,440 three-digit codes. The total number of access codes possible is 180 + 1,440 = 1,620.
EXAMPLE 9: A school committee consists of 8 administrators and 7 teachers. In how many ways can the committee choose a chairperson, a vice-chairperson, and a secretary if the chairperson must be an administrator and the vice-chairperson must be a teacher? There are 8 ways to choose the chairperson and 7 ways to choose the vice-chairperson. The secretary can be any of the 15 committee members except the 2 already chosen. So, there are 13 ways to choose the secretary. By the counting principle, there are 8 × 7 × 13 = 728 ways to choose the three people.
TIP Combinations and permutations do not appear on the Math 1 test. For example, you do not have to know how many ways there would be to choose 3 of the 8 administrators to serve on a subcommittee.