2024-08-23

mdgBookSVG2v1.tex

@@ -175,7 +175,7 @@ \subsection{Table of Measures}\label{tabMeas}
 	\label{fig:meas1}
 \end{figure}
 
-Note that for each of the first five of the six bars of any {\it Einfall} DCSM, there are $9\times 9 = 81$ possible combinations of G- and F-clefs since there are nine choices for each of the G- and F-clefs. However, for the sixth bar (final bar) of each {\it Einfall } DCSM, although there are nine choices for the bars for the G-clef, there is only one choice of a bar for the F-clef, that choice being a middle C of four counts (C4). Thus, there are only nine combinations of G- and F-clefs for the sixth bar (see Figure~\ref{fig:meas2}). All told, these imply that the total number of unique {\it Einfall} DCSMs is $(9^2)^5 \times (9\times 1) = 81\times 81 \times 81 \times 81 \times 81\times 81 \times 9$ = $31,\!381,\!059,\!609$. (Note that this totla number may be doubled by switching the roles of the G- and F-clefs as the DCSMs are invertible at the octave.)
+Note that for each of the first five of the six bars of any {\it Einfall} DCSM, there are $9\times 9 = 81$ possible combinations of G- and F-clefs since there are nine choices for each of the G- and F-clefs. However, for the sixth bar (final bar) of each {\it Einfall } DCSM, although there are nine choices for the bars for the G-clef, there is only one choice of a bar for the F-clef, that choice being a middle C of four counts (C4). Thus, there are only nine combinations of G- and F-clefs for the sixth bar (see Figure~\ref{fig:meas2}). All told, these imply that the total number of unique {\it Einfall} DCSMs is $(9^2)^5 \times (9\times 1) = 81\times 81 \times 81 \times 81 \times 81\times 81 \times 9$ = $31,\!381,\!059,\!609$. (Note that this total number may be doubled by switching the roles of the G- and F-clefs as the DCSMs are invertible at the octave.)
 
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