Counting Rules

The calculation of probability includes counting of the number of favorable outcomes & the total number of outcomes. Though, in many instances, the number of possible outcomes may be so large and 5 which are most commonly used are discussed below:

Counting rule 1

If any one of k various mutually exclusive and collectively exhaustive events can occur on each of n trials, the no. of possible outcomes is equal to Kn. For e.g., if a coin is tossed 5 times, then the number of outcomes is 25 = 2 x 2 x 2 x 2 x 2 = 32.

Counting rule 2

If there are k1 events on the first trial, K2 on the second trial, & kn events on the nth trial, then the number of possible outcomes are

(k1) (k2)...... (kn)

For e.g., if a restaurant menu has a choice of 6 beverages, 3 appetizers, 9 entrees, & 5 desserts, the total number of possible diners would be (3) (6) (9) (5) = 810.

Counting rule 3

The number of ways that all n objects can be placed in order is as follows:

N = n (n - 1) (n - 2)...

Where n is termed as n factorial and 0 is defined as 1.

The number of ways that 8 encyclopedias can be arranged is:

8 = 8 x 7 x  6 x 5 x 4 x 3 x 2 x 1 = 40320.

Counting rule 4

In many illustrations it is important to know the number of ways that r objects selected from n objects (rn) can be arranged in order. For e.g., if in the previous situation 8 encyclopedias are included but there is a room for only 5 on the shelf. In how many ways it can be arranges on the shelf. This can be finding out through the rule of permutations. The no. of ways of arranging r objects selected from n objects in order is given by n / (n - r) 

The number of ordered arrangements of 5 encyclopedias selected from the 8 encyclopedias is equal to:

n/(n - r) = 8/(8 - 5) = 8/3 = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1/3 x 2 x 1 = 6720

Counting rule 5

In many cases we are not interested in the order of the outcomes but only in the number of ways that r objects can be selected out of n objects, irrespective of order. This rule is known as the rule of combinations. The number of ways of selecting r objects out of n objects irrespective of the order is equal to

n/r (n - r)

This expression may be represented by the symbol (n). The number of combinations of 3 journals selected from 5 journals is expressed by (5C3).

This is equal to

n/r (n - r) = 5/3 (5 - 3) = 5/3 x 2 = 5 x 4 x 3 x 2 x 1/3 x 2 x 1 x 2 x 1 = 10.