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Question Number 87731 by mathocean1 last updated on 05/Apr/20
solve in N  A_n ^4 =A_n ^3
$$\mathrm{solve}\:\mathrm{in}\:\mathbb{N} \\ $$$$\mathrm{A}_{\mathrm{n}} ^{\mathrm{4}} =\mathrm{A}_{\mathrm{n}} ^{\mathrm{3}} \\ $$
Commented by Tony Lin last updated on 06/Apr/20
n(n−1)(n−2)(n−3)=n(n−1)(n−2)  n(n−1)(n−2)(n−4)=0  ∵ n≥4  ∴n=4
$${n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right) \\ $$$${n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\because\:{n}\geqslant\mathrm{4} \\ $$$$\therefore{n}=\mathrm{4} \\ $$

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