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1-Given-P-n-1-2n-1-P-n-2n-1-3-5-find-n-2-in-how-many-ways-can-6-persons-stand-in-a-queue-3-how-many-different-4-letter-words-can-be-formed-by-using-letters-of-EDUCATIO

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Question Number 115170 by bobhans last updated on 24/Sep/20
(1)Given ((P _(n−1)^(2n+1) )/(P _n^(2n−1) )) = (3/5) , find n = ? (2) in how many ways can 6 persons stand in a queue? (3) how many different 4 letter words can be formed by using letters of EDUCATION using each letter at most once ?
$$\left(\mathrm{1}\right){Given}\:\frac{{P}\:_{{n}−\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} }{{P}\:_{{n}} ^{\mathrm{2}{n}−\mathrm{1}} }\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:,\:{find}\:{n}\:=\:? \\ $$$$\left(\mathrm{2}\right)\:{in}\:{how}\:{many}\:{ways}\:{can}\:\mathrm{6}\:{persons} \\ $$$${stand}\:{in}\:{a}\:{queue}? \\ $$$$\left(\mathrm{3}\right)\:{how}\:{many}\:{different}\:\mathrm{4}\:{letter}\:{words} \\ $$$${can}\:{be}\:{formed}\:{by}\:{using}\:{letters}\:{of}\: \\ $$$${EDUCATION}\:{using}\:{each}\:{letter}\:{at}\: \\ $$$${most}\:{once}\:? \\ $$$$ \\ $$
Answered by bemath last updated on 24/Sep/20
(1) ((((2n+1)!)/((n+2)!))/(((2n−1)!)/((n−1)!))) = (3/5) → (((2n+1)!(n−1)!)/((2n−1)!(n+2)!)) = (3/5) (((2n+1).2n.(n−1)!)/((n+2)(n+1)n(n−1)!)) = (3/5) ((4n+2)/((n+2)(n+1))) = (3/5) 20n+10 = 3(n^2 +3n+2) 3n^2 −11n−4 = 0 (3n+1)(n−4) = 0 → n=4
$$\left(\mathrm{1}\right)\:\frac{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\left({n}+\mathrm{2}\right)!}}{\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!}}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:\rightarrow\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!\left({n}−\mathrm{1}\right)!}{\left(\mathrm{2}{n}−\mathrm{1}\right)!\left({n}+\mathrm{2}\right)!}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right).\mathrm{2}{n}.\left({n}−\mathrm{1}\right)!}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)!}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\frac{\mathrm{4}{n}+\mathrm{2}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\mathrm{20}{n}+\mathrm{10}\:=\:\mathrm{3}\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\mathrm{3}{n}^{\mathrm{2}} −\mathrm{11}{n}−\mathrm{4}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}{n}+\mathrm{1}\right)\left({n}−\mathrm{4}\right)\:=\:\mathrm{0}\:\rightarrow\:{n}=\mathrm{4} \\ $$
Answered by bemath last updated on 24/Sep/20
(2)^6 C_6 ×6! = 720
$$\left(\mathrm{2}\right)\:^{\mathrm{6}} {C}_{\mathrm{6}} \:×\mathrm{6}!\:=\:\mathrm{720} \\ $$
Answered by bemath last updated on 24/Sep/20
(3)^9 C_4 ×4! = ((9!)/(4!.5!)) ×4! = ((9!)/(5!))
$$\left(\mathrm{3}\right)\:^{\mathrm{9}} {C}_{\mathrm{4}} ×\mathrm{4}!\:=\:\frac{\mathrm{9}!}{\mathrm{4}!.\mathrm{5}!}\:×\mathrm{4}!\:=\:\frac{\mathrm{9}!}{\mathrm{5}!} \\ $$

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