Multiplication Rule If an event can be done in m ways and another event in n ways, then the total number of ways is obtained by multiplying the number of ways these events can be made. Let E 1 = m ways E 2 = n ways TNW(total number of ways) = m x n

Example 1 How many ways can a person enter and go out a 3-door room? Let E 1 = event of entering in the room = 3 ways E 2 = event of going out from the room = 3 ways TNW = (3) (3) = 9 ways

Example 2 A member is allowed to nominate 2 persons among 6 qualified applicants for the position of mayor and vice mayor. I how many ways can he do that? Let E 1 = nomination for mayoral position = 6 ways E 2 = nomination for vice mayoral position = 5 ways TNW = (6) (5) = 30 ways

Notice that the multiplication rule only involves two events. What about if we have more than 2 events? How can we compute the total number of ways in which more than two events can be made? In this type of problem, we use the generalized multiplication rule. Generalized Multiplication Rule If an even 1 (E 1 ) can be done in n 1 of ways, E 2 in n 2 ways and E k in n k ways, then the total number of ways is obtained by TNW = n 1 .n 2 ….n k ways Example 3 How many 4-letter passwords of distinct characters can be formed using the letters a to o? Let E 1 = choice of 1 st letter = 15 ways E 2 = choice of 2 nd letter = 14 ways E 3 = choice of 3 rd letter = 13 ways E 4 = choice of 4 th letter = 12 ways TNW = (15) (14) (13) (12) = 32, 760 4-letter words

Example 4

How many 3-digit numbers can be formed using the digits 1 to 7? a. If the 3-digit number is an odd number? b. If the 3-digit number is an even number? c. If the hundreds digit of the 3-digit number is an odd number and the unit-digit is an even number? d. If the 3-digit number is greater than 400? Note: Repetition of digit is not allowed Solutions: a. Take note that there is a condition here. The 3-digit number must be an odd number. A number is a 3-digit odd number if the unit digit is an odd number. In problems where condition exists, it is recommended to determine firs the number of ways of the event where the condition lies. The solution is shown below. Let E 1 = units digits = 4 ways (1, 3, 5, and 7) E 2 = tens digit = 6 ways (any remaining digits) E 3 = hundreds digit = 5 ways (any of the remaining digits) TNW = (4) (6) (5) = 120 3-digit odd numbers

b. (See explanation above) Let E 1 = units digits = 3 ways (2, 4, and 6) E 2 = tens digit = 6 ways E 3 = hundreds digit = 5 ways TNW = (3) (6) (5) = 90 3-digit even numbers

c. Here, there are two conditions: one, the hundreds digit must be an odd number, and two; units digit must be an even number. The solution is shown below Let E 1 = hundreds digit which must be an odd number = 4 ways E 2 = ones digit which must be an even number = 3 ways E 3 = tens digit will be any of the remaining digits = 5 ways TNW = (3) (4) (5) = 60 3-digit even numbers

d. Again, this problem has a condition. The condition lies on the hundreds digit. The hundreds digit must be a digit greater than or equal to 4. The solution is shown below Let TNW = (4) (6) (5)

= 120 3-digit numbers greater than 400