Counting & Probability
1. Factorial Notation
Factorial notation is used to write the product of all the positive whole numbers up to a given number.
Definition of n!
n factorial is the product of all the integers from 1 to n
"n factorial" is written with an exclamation mark as follows: n!
n! = (n)(n − 1)(n − 2)...
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Number of Outcomes of an Event:
E = an event
n(E) = number of outcomes of event E
If E1 and E2 are mutually exclusive events, and E is the situation where either event E1 or event E2 will occur, then the number of times event E will occur can be given by the expression:
n(E) = n(E1) + n(E2)
If event E1 can result in any one of n(E1) possible outcomes; and for each outcome of the event E1, there are n(E2) possible outcomes of event E2, then there will be n(E1) × n(E2) possible outcomes of the two events.
n(E) = n(E1) × n(E2) (where event E is the event that both E1 and E2 must occur)
How many different combinations can you choose from 2 t-shirts and 4 pairs of jeans?
There are 2 t-shirts and with each t-shirt we could pick 4 pairs of jeans.
2×4=8 possible combinations.
Theorem 1: Arranging n Objects
n distinct objects can be arranged in n! ways.
Theorem 2: Number of Permutations
The number of permutations of n distinct objects taken r at a time, denoted by nPr, where repetitions are not allowed, is: nPr = n(n−1)(n−2)...(n−r+1) = n! ÷ (n−r)!
Example: In how many ways can a supermarket manager display5 brands of cereals in 3 spaces on a shelf?
- 5P3 = 5! ÷ [(5-3)!] = 5! ÷ 2! = 60
Theorem 3: Permutations of Different Kinds of Objects
The number of different permutations of n objects of which n1 are of one kind, n2 are of a second kind, ... nk are of a k-th kind is: n! ÷ (n1! × n2! × n3! ×... nk!)
Theorem 4: Arranging Objects in a Circle
n distinct objects in a circle can be arranged in (n−1)! ways
Number of Combinations:
The number of combinations in which r objects can be selected from a set of n objects, where repetition is not allowed, is denoted by: nCr = n! ÷ [r! (n - r)!]
Suppose an event E can happen in r ways out of a total of n possible equally likely ways.
Then the probability of occurrence of the event is denoted by: P(E) = r ÷ n
The probability of non-occurrence of the event is denoted by: P(É) = 1 - (r ÷ n)
Therefore, P(E) + P(É) = 1 (the sum of probabilities in any experiment is 1)
Using Sample Spaces:
When an experiment is performed, a sample space of all possible outcomes is set up.
In a sample of N equally likely outcomes, a chance of 1N is assigned to each outcome.
The probability of an event for such a sample is the number of outcomes favorable to E divided by the total number of equally likely outcomes in the sample space S of the experiment.
P(E)=n(E) ÷ n(S)
n(E) is the number of outcomes favorable to E
n(S) is the total number of equally likely outcomes in the sample space S of the experiment.
Properties of Probability:
- 0 ≤ P(event) ≤ 1
- P(impossible event) = 0
- P(certain event) = 1
Let E be the event "getting the 3 of diamond". An examination of the sample space shows that there is one "3 of diamond" so that n(E) = 1 and n(S) = 52. Hence the probability of event E occuring is given by P(E) = 1 / 52