91Ó°ÊÓ

The value of ${}^{\mathbf{9}}{\mathbf{C}}_{\mathbf{4}}$ can be computed using the following formula where $\mathbf{N}\mathbf{=}\mathbf{9}$ and $\mathbf{n}\mathbf{=}\mathbf{4}$.

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{9}}{\mathbf{4}}\right)\mathbf{=}\frac{\mathbf{9}\mathbf{!}}{\mathbf{4}\mathbf{!}\left(\mathbf{9}\mathbf{-}\mathbf{4}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{9}\mathbf{×}\mathbf{8}\mathbf{×}\mathbf{7}\mathbf{×}\mathbf{6}\mathbf{×}\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}}{\left(\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)\mathbf{×}\left(\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)} \\ \mathbf{=}\frac{\mathbf{9}\mathbf{×}\mathbf{8}\mathbf{×}\mathbf{7}\mathbf{×}\mathbf{6}}{\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}} \\ \mathbf{=}\mathbf{126} \end{gathered}$$

Thus, the value of${}^{\mathbf{9}}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{126}$.

The value of ${}^{\mathbf{7}}{\mathbf{C}}_{\mathbf{2}}$ can be computed using the following formula where $\mathbf{N}\mathbf{=}\mathbf{7}$ and $\mathbf{n}\mathbf{=}\mathbf{2}$.

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{7}}{\mathbf{2}}\right)\mathbf{=}\frac{\mathbf{7}\mathbf{!}}{\mathbf{2}\mathbf{!}\left(\mathbf{7}\mathbf{-}\mathbf{2}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{7}\mathbf{×}\mathbf{6}\mathbf{×}\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}}{\left(\mathbf{2}\mathbf{×}\mathbf{1}\right)\mathbf{×}\left(\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)} \\ \mathbf{=}\frac{\mathbf{7}\mathbf{×}\mathbf{6}}{\mathbf{2}\mathbf{×}\mathbf{1}} \\ \mathbf{=}\mathbf{21} \end{gathered}$$

Thus, the value of ${}^{\mathbf{7}}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{21}$.

The value of ${}^{\mathbf{4}}{\mathbf{C}}_{\mathbf{4}}$ can be computed using the following formula where $\mathbf{N}\mathbf{=}\mathbf{4}$ and $\mathbf{n}\mathbf{=}\mathbf{4}$.

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{4}}{\mathbf{4}}\right)\mathbf{=}\frac{\mathbf{4}\mathbf{!}}{\mathbf{4}\mathbf{!}\left(\mathbf{4}\mathbf{-}\mathbf{4}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}}{\left(\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)\mathbf{×}\mathbf{0}\mathbf{!}} \\ \mathbf{=}\mathbf{1} \end{gathered}$$

Thus, the value of${}^{\mathbf{4}}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{1}$ .

The value of ${}^{\mathbf{5}}{\mathbf{C}}_{\mathbf{0}}$ can be computed using the following formula where$\mathbf{N}\mathbf{=}\mathbf{5}$ and $\mathbf{n}\mathbf{=}\mathbf{0}$

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{5}}{\mathbf{0}}\right)\mathbf{=}\frac{\mathbf{5}\mathbf{!}}{\mathbf{0}\mathbf{!}\left(\mathbf{5}\mathbf{-}\mathbf{0}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}}{\mathbf{0}\mathbf{!}\mathbf{×}\left(\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)} \\ \mathbf{=}\mathbf{1} \end{gathered}$$

Thus, the value of ${}^{\mathbf{5}}{\mathbf{C}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$

The value of ${}^{\mathbf{6}}{\mathbf{C}}_{\mathbf{5}}$ can be computed using the following formula where $\mathbf{N}\mathbf{=}\mathbf{6}$ and $\mathbf{n}\mathbf{=}\mathbf{5}$

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{6}}{\mathbf{5}}\right)\mathbf{=}\frac{\mathbf{6}\mathbf{!}}{\mathbf{5}\mathbf{!}\left(\mathbf{6}\mathbf{-}\mathbf{5}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{×}\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}}{\left(\mathbf{5}\mathbf{×}\mathbf{4}\mathbf{×}\mathbf{3}\mathbf{×}\mathbf{2}\mathbf{×}\mathbf{1}\right)\mathbf{×}\left(\mathbf{1}\right)} \\ \mathbf{=}\mathbf{6} \end{gathered}$$

Thus, the value of${}^{\mathbf{6}}{\mathbf{C}}_{\mathbf{5}}\mathbf{=}\mathbf{6}$.