Question Number 21799 by Tinkutara last updated on 04/Oct/17
$$\mathrm{There}\:\mathrm{are}\:{n}\:\mathrm{white}\:\mathrm{and}\:{n}\:\mathrm{red}\:\mathrm{balls} \\ $$$$\mathrm{marked}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:….{n}.\:\mathrm{The}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{ways}\:\mathrm{we}\:\mathrm{can}\:\mathrm{arrange}\:\mathrm{these}\:\mathrm{balls}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{row}\:\mathrm{so}\:\mathrm{that}\:\mathrm{neighbouring}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{of} \\ $$$$\mathrm{different}\:\mathrm{colours}\:\mathrm{is} \\ $$
Commented by mrW1 last updated on 06/Oct/17
$$\mathrm{2}\left(\mathrm{n}!\right)^{\mathrm{2}} \:? \\ $$
Commented by Tinkutara last updated on 06/Oct/17
$$\mathrm{Yes}. \\ $$