Category: Permutation and Combination

Question Number 143740 by Willson last updated on 17/Jun/21 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}2n}−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{n}\right)+\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\:\mathrm{ln}\left({e}^{\pi} −{e}^{−\pi} \right) \\ $$ Answered by TheHoneyCat last updated…

Question Number 143627 by bobhans last updated on 16/Jun/21 Answered by Olaf_Thorendsen last updated on 16/Jun/21 $$\mathrm{X}\:=\:\frac{\left(\mathrm{3}^{{x}} −\mathrm{4}^{{x}} \right)\left(\mathrm{3}^{{x}} −\mathrm{4}.\mathrm{4}^{{x}} \right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{6}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{4}}{\mathrm{3}}\right)\right)} \\ $$$${etc}… \\…

Question Number 143462 by Willson last updated on 14/Jun/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{{x}−\mathrm{1}}{{ln}\left(\frac{{x}}{\mathrm{2}−{x}}\right)}\:=\:??? \\ $$ Answered by Dwaipayan Shikari last updated on 14/Jun/21 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}−\mathrm{1}}{{log}\left(\frac{{x}}{\mathrm{2}−{x}}\right)}=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{z}}{{log}\left(\frac{\mathrm{1}+{z}}{\mathrm{1}−{z}}\right)}=\frac{{z}}{\mathrm{2}\left({z}+\frac{{z}^{\mathrm{3}} }{\mathrm{3}}+..\right)}=\frac{\mathrm{1}}{\mathrm{2}}…

Question Number 12087 by tawa last updated on 12/Apr/17 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{10}\:\mathrm{objects}\:\mathrm{be}\:\mathrm{split}\:\mathrm{into}\:\mathrm{two}\:\:\mathrm{groups}\:\mathrm{containing}\: \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{6}\:\mathrm{objects}\:\mathrm{respectively}\:? \\ $$ Answered by sandy_suhendra last updated on 12/Apr/17 $$\mathrm{10C4}\:=\:\frac{\mathrm{10}!}{\mathrm{6}!\:\mathrm{4}!}\:=\mathrm{210} \\ $$$$\mathrm{or}\:\:\mathrm{10C6}\:=\:\mathrm{210} \\…

Question Number 11855 by tawa last updated on 02/Apr/17 $$\mathrm{Ten}\:\mathrm{men}\:\mathrm{are}\:\mathrm{present}\:\mathrm{at}\:\mathrm{a}\:\mathrm{club}.\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{four}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{to} \\ $$$$\mathrm{play}\:\mathrm{bridge}\:\mathrm{if}\:\mathrm{two}\:\mathrm{men}\:\mathrm{refuse}\:\mathrm{to}\:\mathrm{sit}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{table}. \\ $$ Answered by sandy_suhendra last updated on 03/Apr/17 $$\mathrm{let}\:\mathrm{the}\:\mathrm{2}\:\mathrm{men}\:\mathrm{are}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B} \\ $$$$\mathrm{if}\:\mathrm{A}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{and}\:\mathrm{B}\:\mathrm{isn}'\mathrm{t}=\:\mathrm{8C3} \\…