Previous in Permutation and Combination Next in Permutation and Combination
Question Number 131464 by bemath last updated on 05/Feb/21
$$\:\begin{array}{|c|c|}{{old}\:{plate}\::\:{EEU}\:\mathrm{874}}\\{{new}\:{plate}\::\:\mathrm{1}{BXK}\:\mathrm{267}}\\\hline\end{array} \\ $$$${Old}\:{California}\:{license}\:{plate}\: \\ $$$${consisted}\:{of}\:{a}\:{sequence}\:{of} \\ $$$${three}\:{letters}\:{followed}\:{by}\:{three} \\ $$$${digits}\:\left({see}\:{figure}\:{above}\right). \\ $$$${Assuming}\:{that}\:{any}\:{sequence} \\ $$$${of}\:{letters}\:{and}\:{digits}\:{was}\:{allowed} \\ $$$$\left({though}\:{actually}\:{some}\:{combinations}\right. \\ $$$$\left.{of}\:{letters}\:{were}\:{disallowed}\right),\:{how} \\ $$$${many}\:{license}\:{plate}\:{were}\: \\ $$$${available}\:? \\ $$
Answered by mr W last updated on 05/Feb/21
$$\mathrm{26}^{\mathrm{3}} ×\mathrm{10}^{\mathrm{3}} =\mathrm{260}^{\mathrm{3}} =\mathrm{17}\:\mathrm{576}\:\mathrm{000} \\ $$