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Question Number 111272 by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Tricolours}\:\mathrm{flags}\left(\mathrm{each}\:\mathrm{flag}\:\mathrm{having}\right. \\ $$$$\mathrm{three}\:\mathrm{different}\:\mathrm{strips}\:\mathrm{of} \\ $$$$\left.\mathrm{non}−\mathrm{overlapping}\:\mathrm{colours}\right)\:\mathrm{are}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{designed}\:\mathrm{using}\:\mathrm{white},\mathrm{blue},\mathrm{red},\mathrm{yellow} \\ $$$$\mathrm{and}\:\mathrm{black}\:\mathrm{strips}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{flags}\:\mathrm{have}\:\mathrm{blue}\:\mathrm{colour}? \\ $$
Answered by 1549442205PVT last updated on 03/Sep/20
$$\mathrm{Number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{flags}\:\mathrm{have}\:\mathrm{blue}\:\mathrm{colour} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{C}_{\mathrm{4}} ^{\mathrm{2}} =\frac{\mathrm{4}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{6} \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Thanks}. \\ $$