If a procedure can be split into two tasks with \(m\) ways to complete the first an \(n\) the second, then there are \(m \times n\) possible approaches
More generally if there are \(k\) tasks, with \(n_i\) ways for completing task \(i\) then there are \(n_1 \times n_2 \ldots \times n_k\) ways of completing the procedure. \[ \prod_{i=1}^k n_i \] In terms of Sets, if each task is disjoint then the number of approaches can be considered as \[ |A \times B| = |A| \cdot |B| \]
If a procedure can either be completed in \(m\) or \(n\) ways, then it can be completed in \(m + n\) ways.
Generalising to a total \(k\) independent tasks this provides \[ \sum_{i=1}^{k}n_i \] Alternatively, considering each task as belonging to a disjoin set we may also find the cardinality of their union. \[ |A \cup B| = |A| + |B| \]
Also: Inclusion-exclusion principle If a choice can be made between two options, each with \(m\) or \(n\) items. The combined possibility are \(m + n -k\) where \(k\) is the number of items in common.
In set form, we may consider this as a case of intersecting sets. \[ |A \cup B| = |A| + |B| - |A \cap B| \]
If a procedure can be completed in \(n\) ways, and for each way, exactly \(d\) of the \(n\) ways correspond to that way. The total number of ways to complete the procedure is then \(\frac n d\).
This further generalises to \(\frac {n} {d_1 \cdot d_2 \cdot ... d_n}\) where \(d_n\) is the number of repeats of each common element.
Within the set context, if the set \(A\) is the union of \(n\) pair-wise disjoint subsets each with \(d\) elements, then \(n = \frac {|A|} d\).