To calculate the probability, we usually have to count: count the number of elements, count the number of outcomes, ... To count effectively, we can use counting rules.
Fundamental counting rule
A sequence consists of `k` events, the first one has `n_1` possibilities, the second event has `n_2` possibilities, and so on. The total number of possibilities of this sequence is:
`n_1xxn_2xxcdotsxxn_k`
Example 13 : A company would like to produce juice from fruit or vegetable. There are 5 possibilities for the type of fruit or vegetable: orange, pineapple, carrot, mango, and grape, 3 possibilities for the packaging: glass, PET, and aluminum, 3 possibilities for the volume: 250 ml, 330 ml, and 500 ml, 2 possibilities for the type of beverage: ordinary and diet. So the total number of types of juice can be produced is:
5 × 3 × 3 × 2 = 90
Factorial rule
The factorial operation of a positive integer `n`, symbolized as `n!`, is the product of `n` decreasing positive integer from `n` to 1. It mean that:
`n! =nxx(n-1)xx(n - 2)xxcdotsxx3xx2xx1`(8)
To arrange `n` different objects in order, we have `n!` ways. This is the factorial rule.
Example 13 : with 3 characters A, B, and C, we have 3! = 6 ways to combine them. They are ABC, ACB, BAC, BCA, CAB, and CBA.
Permutation rule
A permutation is defined as an arrangement of objects into order. In Example 13 we have 6 permutations from 3 characters.
The permutation rule can be presented as : the number of permutation of `r` objects selected from `n` objects (`n>r`) and without replacement is:
| `P(n,r)=(n!)/((n-r)!)` | (9) |
Example 14 : From a class of 80 students, we would like to choose a representative, an academic associate and an associate for other activities. The number of possibilities is:
`P(80,\ 3)=(80!)/(77!)=80xx79xx78=492.960`
Combination rule
A combination is a group of distinct objects without regard to their order. In Example 13, there are 6 permutation but there is only one combination of 3 characters.
The number of combination of `r` objects selected from `n` objects (`n>r`) and without replacement is:
| `((n),(r))=(n!)/(r!(n-r)!)` | (10) |
Example 15 : From a class of 80 students, we would like to choose 3 students. The number of possibilities is:
`((80),(3))=(80!)/(3!xx77!)=82160`
