Suppose that two experiments (jobs) are to be performed.
Then if experiment 1 can result in any one of p possible outcomes and there are q possible outcomes of experiment 2.
Then we can do either experiment 1 OR experiment 2 (but not both), in p + q ways.

Example If there are 5 odd numbers and 9 even and you must pick one number. How many ways you can do that?
5 + 9 = 14 ways.

## The Multiplication (Product ) Rule

Suppose that two experiments (jobs) are to be performed.
Then if experiment 1 can result in any one of p possible outcomes and if, for each outcome of experiment 1, there are q possible outcomes of experiment 2, then together (AND) there are p * q possible outcomes of two experiments.

Example A small community consists of 10 women, each whom has 3 children. If one woman and one of her children are to be chosen as mother and child of the year, how many different choices are possible?

3 * 10 = 30 ways

## Replacement

1, 2, 3, 4, 5 (a) How many ways can we choose 2 numbers at random (equally likely) from the above 5, with replacement ?

“With replacement” means that after the first number is picked it is “replaced” in the set of numbers, so it could be picked again as the second number.

(a) 5 * 5 = 25 ways

(b) Find the probability that one number is even?

• “The first number is even AND the second is odd, OR, the first is odd AND the second is even.”
• We can then use the addition and multiplication rules to calculate that.

There are
(2 x 3) + (3 x 2) = 12 ways for this event to occur
P (one number is even) = 12 /25

Example How many different 7-place license plates are possible

(a) If the first 3 places are to be occupied by letters and the final 4 by numbers? (with replacement)
26 x 26 x 26 x 10 x 10 x 10 x 10 = 175,760,000

(b) If repetition among letters or numbers were prohibited? (without replacement)
26 x 25 x 24 x 10 x 9 x 8 x 7 = 78624000

## Permutations

A permutation, also called an “arrangement number” or “order,” are arrangement of the elements of an ordered list.

Suppose that we have n distinct objects, then the number of ways to arrange these n objects in a row (using each symbol once and only once) is
n! = n (n-1) (n-2)…1

Special definition: 0!=1

Suppose the letters (a, b, c, d, e, f) are arranged at random to form a six-letter word (an arrangement) – we must use each letter once only (.?….)

(a) What is the sample space?
S = { abcdef, abcdfe,…, fedcba }
6!=720
Or using the basic principle of counting 6 x 5 x 4 x 3 x 2 x 1 = 720

Question Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language book. Ms. Jones wants to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible?

4! 4! 3! 2! 1! = 6912