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Question Number 16868 by tawa tawa last updated on 27/Jun/17

In how many ways can the letters of the word. EVERMORE be arrange  if the word must begin with   (i) R  (ii) E

$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}.\:\mathrm{EVERMORE}\:\mathrm{be}\:\mathrm{arrange} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{word}\:\mathrm{must}\:\mathrm{begin}\:\mathrm{with}\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{R} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{E} \\ $$

Answered by mrW1 last updated on 27/Jun/17

(i) R begins  ((7!)/(3!))=840  (ii) E begins  ((7!)/(2!2!))=1260    (iii) no restriction  ((8!)/(3!2!))=3360

$$\left(\mathrm{i}\right)\:\mathrm{R}\:\mathrm{begins} \\ $$$$\frac{\mathrm{7}!}{\mathrm{3}!}=\mathrm{840} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{E}\:\mathrm{begins} \\ $$$$\frac{\mathrm{7}!}{\mathrm{2}!\mathrm{2}!}=\mathrm{1260} \\ $$$$ \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{no}\:\mathrm{restriction} \\ $$$$\frac{\mathrm{8}!}{\mathrm{3}!\mathrm{2}!}=\mathrm{3360} \\ $$

Commented by tawa tawa last updated on 27/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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