Counting applies to probabilistic experiments where we have a finite number of outcomes and each outocme has the same probability of occuring.

Basic counting principle

Suppose in an experiment, we have multiple stages. If in stage 1 we have \(m_1\) outcomes, in stage 2 we have \(m_2\) outcomes and in stage 3 we have \(m_3\) outcomes, then the total number of possible outcomes for the experiment are \(m_1 \times m_2 \times m_3\)

Example: How many license plates can we make if the first \(3\) positions are alphabets and the last \(4\) positions are digits?

Solution:

Since the first three positions are alphabets, we have \(26\) choices for the first position, \(26\) choices for the second position and \(26\) choices for the third position. Since the last four positions are digits, we have \(10\) choices for the fourth position, \(10\) choices for the fifth position, \(10\) choices for the sixth position and \(10\) choices for the seventh position.

Therefore in total we have \(26 \times 26 \times 26 \times 10 \times 10 \times 10 \times 10\) number of license plates possible.

Now if repitation is not allowed then in total we have \(26 \times 25 \times 24 \times 10 \times 9 \times 8 \times 7\) number of license plates possible.

[Permutation]

[Combination]