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pt100/pt100.tex

@@ -620,7 +620,7 @@ We can use the equation \ref{eqn:correctedccps2}, reproduced below to verify thi
 \begin{equation}
  |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}   
 - \sum^{p}_{2..cc}{{\sum^{j}_{j \in J}   \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}}  }
- \label{eqn:correctedccps2}
+ %\label{eqn:correctedccps2}
 \end{equation}
 
 
@@ -629,12 +629,13 @@ $|FM(C_j)|$ is always 2 here,  as all the components are resistors and have two
 %
 % Factorial of zero is one ! You can only arrange an empty set one way !
 
-Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ is always 2 here as all the components are resistors and have two failure modes.
+Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ = 2.
+%is always 2 for this circuit, as all the components are resistors and have two failure modes.
 
 \begin{equation}
  |{\mathcal{P}_{2}SU}| = {\sum^{k}_{1..2}  \frac{6!}{k!(6 - k)!}}   
 - \sum^{p}_{2..2}{{\sum^{j}_{1..3}   \frac{2!}{p!(2 - p)!}}  }
- \label{eqn:correctedccps2}
+ %\label{eqn:correctedccps2}
 \end{equation}
 
 $|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check