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Question Number 133314 by bramlexs22 last updated on 21/Feb/21

$$\mathrm{How}\mathrm{many}\mathrm{rearrangements}\mathrm{are}$$$$\mathrm{there}\mathrm{of}\mathrm{the}\mathrm{letters}\mathrm{in}\mathrm{the}\mathrm{world}$$$$\left(\mathrm{i}\right)\mathrm{ENGINEERING}$$$$\left(\mathrm{ii}\right)\mathrm{MATHEMATICAL}$$

Answered by EDWIN88 last updated on 21/Feb/21

$$\left(\mathrm{i}\right)=\frac{11!}{3!3!2!2!}=277200$$$$\left(\mathrm{ii}\right)=\frac{12!}{2!3!2!}=1.99584\mathrm{E}7=19958400$$

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