91Ó°ÊÓ

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\) ways and the second element in \({{\rm{n}}_{\rm{2}}}\) ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\).

The first element is a Beethoven symphony, for which \({{\rm{n}}_{\rm{1}}}{\rm{ = 9}}\), while the second element is a Beethoven piano concerto, for which \({{\rm{n}}_{\rm{2}}}{\rm{ = 5}}\)

\({\rm{9*5 = 45}}\)

For k-Tuples, there is a Product Rule.

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\) ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

Similarly, there exist potential \({\rm{k}}\)-tuples for an ordered collection of \({\rm{k}}\)components, where the \({{\rm{k}}^{{\rm{th\;}}}}\)can be picked in \({{\rm{n}}_{\rm{k}}}\)ways.

Using this, the first element is the Beethoven symphony (\({{\rm{n}}_{\rm{1}}}{\rm{ = 9}}\)), the second element is the Beethoven piano concerto (\({{\rm{n}}_{\rm{2}}}{\rm{ = 5}}\)), and the third element is the Beethoven piano sonata \({{\rm{n}}_{\rm{3}}}{\rm{ = 32}}\), giving a total of \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\) potential sequences. As a result, this strategy can be carried out for \({\rm{1440}}\) nights in a row, or nearly \({\rm{4}}\) years, without repeating the same programme.