Question Number 85322 by mathocean1 last updated on 20/Mar/20
$${please}\:{help}\:{me}\:{to}\:{solve}\:{this}\:{in}\:\mathbb{N} \\ $$$${A}_{{n}} ^{\mathrm{2}} −\mathrm{3}{C}_{{n}} ^{\:{n}−\mathrm{2}} +{n}=−\mathrm{20} \\ $$
Commented by mr W last updated on 20/Mar/20
$${what}\:{do}\:{you}\:{want}\:{to}\:{get}\:{exactly}? \\ $$$$ \\ $$$${do}\:{you}\:{want}\:{to}\:{get}\:{A}_{{n}} ?\:{this}\:{is} \\ $$$${possible}:\:{A}_{{n}} =\pm\sqrt{\mathrm{3}{C}_{{n}} ^{\:{n}−\mathrm{2}} −{n}−\mathrm{20}} \\ $$$$ \\ $$$${or}\:{do}\:{you}\:{want}\:{to}\:{solve}\:{for}\:{n}?\:{this} \\ $$$${is}\:{not}\:{possible},\:{because}\:{it}\:{is}\:{not}\: \\ $$$${given}\:{what}\:{A}_{{n}} \:{is}. \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${A}_{{n}} ^{\mathrm{2}} =\frac{\mathrm{n}!}{\left({n}−\mathrm{2}\right)!}\:\:{this}\:{is}\:{the}\:{formula}\:{that}\: \\ $$$${i}\:{have}. \\ $$
Commented by mr W last updated on 20/Mar/20
$${what}\:{do}\:{you}\:{mean}\:{with}\:{C}_{{n}} ^{\:{n}−\mathrm{2}} ? \\ $$
Commented by mr W last updated on 20/Mar/20
$${for}\:{combination}\:{calculation} \\ $$$${C}_{{r}} ^{{n}} \:{is}\:{defined}\:{only}\:{for}\:\mathrm{0}\leqslant{r}\leqslant{n} \\ $$$${therefore}\:{C}_{{n}} ^{{n}−\mathrm{2}} \:{is}\:{not}\:{possible}! \\ $$$${or}\:{do}\:{you}\:{mean}\:\left({C}_{{n}} \right)^{{n}−\mathrm{2}} ?\:{then}\:{you} \\ $$$${must}\:{specify}\:{what}\:{is}\:{C}_{{n}} . \\ $$
Commented by mr W last updated on 20/Mar/20
$${i}\:{think}\:{your}\:{question}\:{is}\:{not}\:{correct}. \\ $$$${please}\:{recheck}! \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${I}\:{rechecked}\:{it}\:{sir}... \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${It}\:{is}\:{like}\:{this}. \\ $$$${I}\:{think}\:{that}\:{C}_{{n}} ^{{n}−\mathrm{2}} =\frac{{n}!}{\mathrm{2}×\left({n}−\mathrm{2}\right)!} \\ $$
Commented by mr W last updated on 20/Mar/20
$${you}\:{write}\:{C}_{{n}} ^{{n}−\mathrm{2}} ,\:{but}\:{you}\:{mean}\:{C}_{{n}−\mathrm{2}} ^{{n}} . \\ $$
Commented by mr W last updated on 20/Mar/20
$${so}\:{your}\:{question}\:{is}\:{to}\:{solve}\:{n}\:{for} \\ $$$${A}_{{n}} ^{\mathrm{2}} −\mathrm{3}{C}_{{n}−\mathrm{2}} ^{{n}} +{n}=−\mathrm{20} \\ $$$${with}\:{A}_{{n}} ^{\mathrm{2}} =\frac{{n}!}{\left({n}−\mathrm{2}\right)!} \\ $$$$ \\ $$$$\frac{{n}!}{\left({n}−\mathrm{2}\right)!}−\mathrm{3}×\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}+{n}=−\mathrm{20} \\ $$$$\frac{{n}!}{\left({n}−\mathrm{2}\right)!}−{n}=\mathrm{20} \\ $$$${n}\left({n}−\mathrm{1}\right)−{n}=\mathrm{20} \\ $$$${n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{20}=\mathrm{0} \\ $$$${n}=\mathrm{1}+\sqrt{\mathrm{21}}\:\notin\mathbb{N} \\ $$$$\Rightarrow{no}\:{solution}! \\ $$$$ \\ $$$${again}:\:{the}\:{question}\:{is}\:{wrong}! \\ $$
Commented by mathocean1 last updated on 20/Mar/20
$${yes} \\ $$