Question Number 131162 by EDWIN88 last updated on 02/Feb/21 $${Five}\:{children}\:{sitting}\:{one}\:{behind} \\ $$$${the}\:{other}\:{in}\:{a}\:{five}\:{seater}\:{merry}−{go}−{round} \\ $$$$,{decide}\:{to}\:{switch}\:{seats}\:{so}\:{that}\:{each} \\ $$$${child}\:{has}\:{new}\:{companion}\:{in}\:{front}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{this}\:{be}\:{done}? \\ $$ Commented by mr W last…
Question Number 66 by newuser last updated on 15/Nov/14 $$\mathrm{How}\:\mathrm{many}\:\mathrm{unique}\:\mathrm{arrangements}\:\mathrm{are}\: \\ $$$$\mathrm{possible}\:\mathrm{for}\:\mathrm{a}\:\mathrm{2}×\mathrm{2}\:\mathrm{Rubic}\:\mathrm{cube}? \\ $$ Commented by 123456 last updated on 14/Dec/14 $$\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$ Commented…
Question Number 23 by user1 last updated on 25/Jan/15 $$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\frac{{x}^{{n}} }{{n}!}\:,\:\mathrm{where}\:\boldsymbol{\mathrm{P}}_{{r}} \left({n}\right)\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree}\:{r}\:\mathrm{in}\:{n}. \\ $$ Answered by user1 last updated on…
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 12590 by tawa last updated on 26/Apr/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{basic}\:\mathrm{version}\:\mathrm{of}\:\mathrm{pocker}\:,\:\mathrm{each}\:\mathrm{players}\:\mathrm{is}\:\mathrm{dealth}\:\mathrm{5}\:\mathrm{cards}\:\mathrm{from}\:\mathrm{a}\:\mathrm{standard} \\ $$$$\mathrm{52}\:\mathrm{cards}\:\left(\mathrm{no}\:\mathrm{pocker}\right).\:\mathrm{How}\:\mathrm{many}\:\mathrm{diferent}\:\mathrm{5}\:\mathrm{cards}\:\mathrm{pocker}\:\mathrm{hand}\:\mathrm{are}\:\mathrm{there}\:??? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143627 by bobhans last updated on 16/Jun/21 Answered by Olaf_Thorendsen last updated on 16/Jun/21 $$\mathrm{X}\:=\:\frac{\left(\mathrm{3}^{{x}} −\mathrm{4}^{{x}} \right)\left(\mathrm{3}^{{x}} −\mathrm{4}.\mathrm{4}^{{x}} \right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{6}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{4}}{\mathrm{3}}\right)\right)} \\ $$$${etc}… \\…
Question Number 78083 by TawaTawa last updated on 14/Jan/20 Commented by john santu last updated on 14/Jan/20 $${sir}\:{this}\:{ambigue}\:\mathrm{21600}!\: \\ $$ Commented by TawaTawa last updated…
Question Number 143462 by Willson last updated on 14/Jun/21 $$\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 12218 by tawa last updated on 16/Apr/17 $$\mathrm{5}!!\:=\:? \\ $$ Commented by tawa last updated on 16/Apr/17 $$\mathrm{How}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{15} \\ $$ Commented by prakash…
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} \\…