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Question-114665




Question Number 114665 by mohammad17 last updated on 20/Sep/20
Answered by Olaf last updated on 20/Sep/20
−For the first digit we have 5 solutions  (1, 2, 3, 4 or 5).  −For the second digit we have 4 solutions  (no repetition)  −For the third digit we have 3 solutions  5×4×3 = A_5 ^3  = ((5!)/(2!)) = 60
$$−\mathrm{For}\:\mathrm{the}\:\mathrm{first}\:\mathrm{digit}\:\mathrm{we}\:\mathrm{have}\:\mathrm{5}\:\mathrm{solutions} \\ $$$$\left(\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{or}\:\mathrm{5}\right). \\ $$$$−\mathrm{For}\:\mathrm{the}\:\mathrm{second}\:\mathrm{digit}\:\mathrm{we}\:\mathrm{have}\:\mathrm{4}\:\mathrm{solutions} \\ $$$$\left(\mathrm{no}\:\mathrm{repetition}\right) \\ $$$$−\mathrm{For}\:\mathrm{the}\:\mathrm{third}\:\mathrm{digit}\:\mathrm{we}\:\mathrm{have}\:\mathrm{3}\:\mathrm{solutions} \\ $$$$\mathrm{5}×\mathrm{4}×\mathrm{3}\:=\:\mathrm{A}_{\mathrm{5}} ^{\mathrm{3}} \:=\:\frac{\mathrm{5}!}{\mathrm{2}!}\:=\:\mathrm{60} \\ $$$$ \\ $$$$ \\ $$

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