using 6 different flags,how many different signal can be made by using atleast three flags arranging one above the other?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Total} \mathrm{flags}=6 \\ \mathrm{We} \mathrm{have} \mathrm{four} \mathrm{cases} \mathrm{possible}: \\ \mathrm{Signal} \mathrm{with} \left(3 \mathrm{flags} \mathrm{OR} 4 \mathrm{flags} \mathrm{OR} 5 \mathrm{flags} \mathrm{OR} 6 \mathrm{flags}\right) \\ \Rightarrow \mathrm{Number} \mathrm{of} \mathrm{ways}={ \\ }^6{\mathrm{C}}_3.3!{+}^6{\mathrm{C}}_4.4!{+}^6{\mathrm{C}}_5.5!{+}^6{\mathrm{C}}_6.6! \\ \mathrm{Note} : { }^6{\mathrm{C}}_3.3! \mathrm{means} \mathrm{we} \mathrm{selected} 3 \mathrm{flags} \mathrm{out} \mathrm{of} 6 \mathrm{in}{ }^6{\mathrm{C}}_3 \mathrm{ways} \mathrm{and} \\ \mathrm{then} \mathrm{arranged} \mathrm{them} \mathrm{in} 3! \mathrm{ways}. \\ \Rightarrow \mathrm{Number} \mathrm{of} \mathrm{ways}=\frac{6.5.4.}{3.2.1}.6+\frac{6.5}{2.1}.24+\frac{6}{1}.120+1.720 \\ \mathbf{\Rightarrow }\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ways}\mathbf{=}\mathbf{1920}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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