Question

Compute the number of ways can select n element from N element of each of the following:

1. $\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{5}$
2. $\mathbf{n}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{6}$
3. $\mathbf{n}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{20}$

Short answer

1. 10
2. 20
3. 15504

Step-by-step solution

Computing the number of ways to select 2 from 5

The value of ${}^{\mathbf{5}}{\mathbf{C}}_{\mathbf{2}}$ can be computed using the formula mentioned below:

Here, $\mathbf{N}\mathbf{=}\mathbf{5}$ and $\mathbf{n}\mathbf{=}\mathbf{2}$.

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{5}}{\mathbf{2}}\right)\mathbf{=}\frac{\mathbf{5}\mathbf{!}}{\mathbf{2}\mathbf{!}\left(\mathbf{5}\mathbf{-}\mathbf{2}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\left(\mathbf{2}\mathbf{\times }\mathbf{1}\right)\mathbf{\times }\left(\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)} \\ \mathbf{=}\frac{\mathbf{5}\mathbf{\times }\mathbf{4}}{\mathbf{2}\mathbf{\times }\mathbf{1}} \\ \mathbf{=}\mathbf{10} \end{gathered}$$

Therefore, the value of the number of ways to select2 from 5 using combination is 10.

Computing the number of ways to select 3 from 6

The value of ${}^{\mathbf{6}}{\mathbf{C}}_{\mathbf{3}}$ can be computed using the formula where $\mathbf{N}\mathbf{=}\mathbf{6}$ and$\mathbf{n}\mathbf{=}\mathbf{3}$ .

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{6}}{\mathbf{3}}\right)\mathbf{=}\frac{\mathbf{6}\mathbf{!}}{\mathbf{3}\mathbf{!}\left(\mathbf{6}\mathbf{-}\mathbf{3}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\left(\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)\mathbf{\times }\left(\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)} \\ \mathbf{=}\frac{\mathbf{6}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{4}}{\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}} \\ \mathbf{=}\mathbf{20} \end{gathered}$$

Therefore, the value of${}^{\mathbf{6}}{\mathbf{C}}_{\mathbf{3}}\mathbf{=}\mathbf{20}$.

Computing the number of ways to select 5 from 20

The value of${}^{\mathbf{20}}{\mathbf{C}}_{\mathbf{5}}$can be computed using the formula where$\mathbf{N}\mathbf{=}\mathbf{20}$and$\mathbf{n}\mathbf{=}\mathbf{5}$.

$$\begin{gathered} \left(\frac{\mathbf{N}}{\mathbf{N}}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{\mathbf{n}\mathbf{!}\left(\mathbf{N}\mathbf{-}\mathbf{n}\right)\mathbf{!}} \\ \left(\frac{\mathbf{20}}{\mathbf{5}}\right)\mathbf{=}\frac{\mathbf{20}\mathbf{!}}{\mathbf{5}\mathbf{!}\left(\mathbf{20}\mathbf{-}\mathbf{5}\right)\mathbf{!}} \\ \mathbf{=}\frac{\mathbf{20}\mathbf{\times }\mathbf{19}\mathbf{\times }\mathbf{18}\mathbf{\times }\mathbf{17}\mathbf{\times }\mathbf{16}\mathbf{\times }\mathbf{15}\mathbf{\times }\mathbf{14}\mathbf{\times }\mathbf{13}\mathbf{\times }\mathbf{12}\mathbf{\times }\mathbf{11}\mathbf{\times }\mathbf{10}\mathbf{\times }\mathbf{9}\mathbf{\times }\mathbf{8}\mathbf{\times }\mathbf{7}\mathbf{\times }\mathbf{6}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\left(\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)\mathbf{\times }\left(\mathbf{15}\mathbf{\times }\mathbf{14}\mathbf{\times }\mathbf{13}\mathbf{\times }\mathbf{12}\mathbf{\times }\mathbf{11}\mathbf{\times }\mathbf{10}\mathbf{\times }\mathbf{9}\mathbf{\times }\mathbf{8}\mathbf{\times }\mathbf{7}\mathbf{\times }\mathbf{6}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)} \\ \mathbf{=}\frac{\mathbf{20}\mathbf{\times }\mathbf{19}\mathbf{\times }\mathbf{18}\mathbf{\times }\mathbf{17}\mathbf{\times }\mathbf{16}}{\left(\mathbf{5}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}\right)} \\ \mathbf{=}\mathbf{15},\mathbf{504} \end{gathered}$$

Hence, the value of${}^{\mathbf{20}}{\mathbf{C}}_{\mathbf{5}}\mathbf{=}\mathbf{15504}$.