the weak scattering of quarks, composing the hadron, depends on the quark contents of baryons and results in: $$\begin{aligned}
{\cal T}_{6,PI}^{(1)} &=& 2 {\cal T}_{PI,u\bar d}^c \\
{\cal T}_{6,WS}^{(2)} &=& 2 {\cal T}_{WS,cd} \\
{\cal T}_{6,PI}^{(3)} &=& 2 {\cal T}_{PI,u\bar d}^{c'} + 2 \sum_{l} {\cal
T}_{PI,\nu_l \bar l}^c  \\
{\cal T}_{6,PI}^{(4)} &=& {\cal T}_{PI,u\bar d}^c + {\cal T}_{PI,s\bar c}^b +
{\cal T}_{PI,d\bar u}^b + \sum_l{\cal T}_{PI,l\bar\nu_l}^b\\
{\cal T}_{6,WS}^{(4)} &=& {\cal T}_{WS,bu} + {\cal T}_{WS,bc}\\
{\cal T}_{6,PI}^{(5)} &=& {\cal T}_{PI,s\bar c}^b + {\cal T}_{PI,d\bar u}^b +
{\cal T}_{PI,d\bar u}^{'b} + \sum_l{\cal T}_{PI,l\bar\nu_l}^b\\
{\cal T}_{6,WS}^{(5)} &=& {\cal T}_{WS,cd} + {\cal T}_{WS,bc} \\
{\cal T}_{6,PI}^{(6)} &=& {\cal T}_{PI,u\bar d}^{c'} + \sum_{l} {\cal
T}_{PI,\nu_l \bar l}^c + {\cal T}_{PI,s\bar c}^b + {\cal T}_{PI,d\bar u}^b +
\sum_l{\cal T}_{PI,l\bar\nu_l}^b + {\cal T}_{PI,s\bar c}^{'b}\\
{\cal T}_{6,WS}^{(6)} &=& {\cal T}_{WS,bc} + {\cal T}_{WS,cs}\\
{\cal T}_{6,WS}^{(7)} &=& 2 {\cal T}_{WS,bu} \\
{\cal T}_{6,PI}^{(8)} &=& 2 {\cal T}_{PI,d\bar u}^{'b} \\
{\cal T}_{6,PI}^{(9)} &=& 2 {\cal T}_{PI,s\bar c}^{'b}\end{aligned}$$ so that $$\begin{aligned}
{\cal T}_{PI,s\bar c}^b &=&
-\frac{G_F^2|V_{cb}|^2}{4\pi}m_b^2(1-\frac{m_c}{m_b})^2\nonumber\\
&& ([(\frac{(1-z_{-})^2}{2}- \frac{(1-z_{-})^3}{4})
(\bar b_i\gamma_{\alpha}(1-\gamma_5)b_i)(\bar
c_j\gamma^{\alpha}(1-\gamma_5)c_j) + \nonumber\\
&& (\frac{(1-z_{-})^2}{2} -
\frac{(1-z_{-})^3}{3})(\bar b_i\gamma_{\alpha}\gamma_5 b_i)(\bar
c_j\gamma^{\alpha}(1-\gamma_5)c_j)] \label{16}
\\&& [(C_{+} - C_{-})^2 +
\frac{1}{3}(1-k^{\frac{1}{2}})(5C_{+}^2+C_{-}^2+6C_{-}C_{+})]+ \nonumber\\
&& [(\frac{(1-z_{-})^2}{2} - \frac{(1-z_{-})^3}{4})(\bar
b_i\gamma_{\alpha}(1-\gamma_5)b_j)(\bar c_j\gamma^{\alpha}(1-\gamma_5)c_i) +
\nonumber\\
&&  (\frac{(1-z_{-})^2}{2} - \frac{(1-z_{-})^3}{3})(\bar
