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
$$\frac{\mathrm{11}!}{\mathrm{5}!×\mathrm{2}!×\mathrm{2}!}=\mathrm{83160} \\ $$