Question Number 131464 by bemath last updated on 05/Feb/21
![determinant (((old plate : EEU 874)),((new plate : 1BXK 267))) Old California license plate consisted of a sequence of three letters followed by three digits (see figure above). Assuming that any sequence of letters and digits was allowed (though actually some combinations of letters were disallowed), how many license plate were available ?](https://www.tinkutara.com/question/Q131464.png)
$$\:\begin{array}{|c|c|}{{old}\:{plate}\::\:{EEU}\:\mathrm{874}}\\{{new}\:{plate}\::\:\mathrm{1}{BXK}\:\mathrm{267}}\\\hline\end{array} \\ $$$${Old}\:{California}\:{license}\:{plate}\: \\ $$$${consisted}\:{of}\:{a}\:{sequence}\:{of} \\ $$$${three}\:{letters}\:{followed}\:{by}\:{three} \\ $$$${digits}\:\left({see}\:{figure}\:{above}\right). \\ $$$${Assuming}\:{that}\:{any}\:{sequence} \\ $$$${of}\:{letters}\:{and}\:{digits}\:{was}\:{allowed} \\ $$$$\left({though}\:{actually}\:{some}\:{combinations}\right. \\ $$$$\left.{of}\:{letters}\:{were}\:{disallowed}\right),\:{how} \\ $$$${many}\:{license}\:{plate}\:{were}\: \\ $$$${available}\:? \\ $$
Answered by mr W last updated on 05/Feb/21
![26^3 ×10^3 =260^3 =17 576 000](https://www.tinkutara.com/question/Q131474.png)
$$\mathrm{26}^{\mathrm{3}} ×\mathrm{10}^{\mathrm{3}} =\mathrm{260}^{\mathrm{3}} =\mathrm{17}\:\mathrm{576}\:\mathrm{000} \\ $$