If-n-C-3-n-C-2-14-Find-n-P-2-

Question Number 150289 by mathdanisur last updated on 10/Aug/21

$$\mathrm{If}\:\:\:_{\boldsymbol{\mathrm{n}}} \mathrm{C}_{\mathrm{3}} \:−\:_{\boldsymbol{\mathrm{n}}} \mathrm{C}_{\mathrm{2}} \:=\:\mathrm{14} \\ $$$$\mathrm{Find}\:\:\:_{\boldsymbol{\mathrm{n}}} \mathrm{P}_{\mathrm{2}} \:=\:? \\ $$

Answered by Ar Brandon last updated on 12/Aug/21

^n C_3 − ^n C_2 =14 ((n!)/((n−3)!3!))−((n!)/((n−2)!2!))=14 ((n!(n−2−3))/((n−2)!3!))=((n(n−1)(n−5))/(3!))=14 n^3 −6n^2 +5n−84=0, n=7 ^7 P_2 =((7!)/(5!))=42

$$\overset{{n}} {\:}\mathrm{C}_{\mathrm{3}} −\overset{{n}} {\:}\mathrm{C}_{\mathrm{2}} =\mathrm{14} \\ $$$$\frac{{n}!}{\left({n}−\mathrm{3}\right)!\mathrm{3}!}−\frac{{n}!}{\left({n}−\mathrm{2}\right)!\mathrm{2}!}=\mathrm{14} \\ $$$$\frac{{n}!\left({n}−\mathrm{2}−\mathrm{3}\right)}{\left({n}−\mathrm{2}\right)!\mathrm{3}!}=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{5}\right)}{\mathrm{3}!}=\mathrm{14} \\ $$$${n}^{\mathrm{3}} −\mathrm{6}{n}^{\mathrm{2}} +\mathrm{5}{n}−\mathrm{84}=\mathrm{0},\:{n}=\mathrm{7} \\ $$$$\overset{\mathrm{7}} {\:}\mathrm{P}_{\mathrm{2}} =\frac{\mathrm{7}!}{\mathrm{5}!}=\mathrm{42} \\ $$

Commented by mathdanisur last updated on 10/Aug/21