Question-131994

Question Number 131994 by liberty last updated on 10/Feb/21

Commented by liberty last updated on 10/Feb/21

$$\mathrm{old}\:\mathrm{unswered} \\ $$

Answered by EDWIN88 last updated on 10/Feb/21

(1) n(S) = ((15!)/(5!.5!.5!)) ; n(A) = 3!×((12!)/(4!.4!.4!)) P(A) = ((6×12!×5!×5!×5!)/(4!×4!×4!×15!))=((25)/(91))

$$\left(\mathrm{1}\right)\:\mathrm{n}\left(\mathrm{S}\right)\:=\:\frac{\mathrm{15}!}{\mathrm{5}!.\mathrm{5}!.\mathrm{5}!}\:;\:\mathrm{n}\left(\mathrm{A}\right)\:=\:\mathrm{3}!×\frac{\mathrm{12}!}{\mathrm{4}!.\mathrm{4}!.\mathrm{4}!} \\ $$$$\:\:\:\:\:\:\:\mathrm{P}\left(\mathrm{A}\right)\:=\:\frac{\mathrm{6}×\mathrm{12}!×\mathrm{5}!×\mathrm{5}!×\mathrm{5}!}{\mathrm{4}!×\mathrm{4}!×\mathrm{4}!×\mathrm{15}!}=\frac{\mathrm{25}}{\mathrm{91}} \\ $$

Commented by liberty last updated on 10/Feb/21

$$\:\:\:\mathrm{Vielen}\:\mathrm{Dank}\:\mathrm{Sir}\:\mathrm{Edwin} \\ $$

Answered by EDWIN88 last updated on 10/Feb/21

(2) n(A)= 3×C_2 ^(12) ×C_5 ^(10) ×C_5 ^5 = ((3×12!)/(2!5!.5!)) P(A)= ((3×12!×5!×5!×5!)/(2×5!×5!×15!))= ((3×12!×120)/(2×15×14×13×12!)) = ((3×60)/(15×14×13)) = ((12)/(14×13))=(6/(91))

$$\left(\mathrm{2}\right)\:\mathrm{n}\left(\mathrm{A}\right)=\:\mathrm{3}×\mathrm{C}_{\mathrm{2}} ^{\mathrm{12}} ×\mathrm{C}_{\mathrm{5}} ^{\mathrm{10}} ×\mathrm{C}_{\mathrm{5}} ^{\mathrm{5}} =\:\frac{\mathrm{3}×\mathrm{12}!}{\mathrm{2}!\mathrm{5}!.\mathrm{5}!} \\ $$$$\:\:\:\:\:\:\:\mathrm{P}\left(\mathrm{A}\right)=\:\frac{\mathrm{3}×\mathrm{12}!×\mathrm{5}!×\mathrm{5}!×\mathrm{5}!}{\mathrm{2}×\mathrm{5}!×\mathrm{5}!×\mathrm{15}!}=\:\frac{\mathrm{3}×\mathrm{12}!×\mathrm{120}}{\mathrm{2}×\mathrm{15}×\mathrm{14}×\mathrm{13}×\mathrm{12}!} \\ $$$$\:=\:\frac{\mathrm{3}×\mathrm{60}}{\mathrm{15}×\mathrm{14}×\mathrm{13}}\:=\:\frac{\mathrm{12}}{\mathrm{14}×\mathrm{13}}=\frac{\mathrm{6}}{\mathrm{91}} \\ $$

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