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Question Number 188653 by Shrinava last updated on 04/Mar/23

$$\mathrm{In}\mathrm{how}\mathrm{many}\mathrm{different}\mathrm{ways}\mathrm{can}\mathrm{the}$$$$\mathrm{letters}\mathrm{of}\mathrm{the}\mathrm{word}\mathrm{ABRAKADABRA}$$$$\mathrm{be}\mathrm{arranged}?$$

Commented by Ajetunmobi last updated on 04/Mar/23

$$\frac{11!}{5!\times 2!\times 2!}=83160$$

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