Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

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!))

$$\left(\mathrm{3}\right)\:^{\mathrm{9}} {C}_{\mathrm{4}} ×\mathrm{4}!\:=\:\frac{\mathrm{9}!}{\mathrm{4}!.\mathrm{5}!}\:×\mathrm{4}!\:=\:\frac{\mathrm{9}!}{\mathrm{5}!} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com