Question Number 58373 by Tawa1 last updated on 22/Apr/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{\mathrm{6}} \:\:\mathrm{in}\:\:\:\left(\mathrm{2x}\:+\:\mathrm{1}\right)^{\mathrm{6}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$ Commented by maxmathsup by imad last updated on 24/Apr/19 $${we}\:{have}\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{6}}…
Question Number 123878 by liberty last updated on 29/Nov/20 $${Nine}\:{chairs}\:{in}\:{a}\:{row}\:{are}\:{to}\:{be}\:{occupied}\:{by} \\ $$$${six}\:{students}\:{and}\:{Prof}\:{George}\:,\:{Prof}\:{Pieter}, \\ $$$${and}\:{Prof}\:{John}.\:{These}\:{three}\:{professors}\:{arrive} \\ $$$${before}\:{the}\:{six}\:{students}\:{and}\:{decide}\:{to}\:{choose} \\ $$$${their}\:{chairs}\:{so}\:{that}\:{each}\:{professor}\:{will}\:{be}\:{between} \\ $$$${two}\:{students}.\:{In}\:{how}\:{many}\:{ways}\:{can}\:{Professor}\:{George} \\ $$$$,\:{Pieter}\:{and}\:{John}\:{choose}\:{their}\:{chairs}\:?\: \\ $$ Answered…
Question Number 123695 by pipin last updated on 27/Nov/20 $$\boldsymbol{\mathrm{There}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{ways}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{answer}} \\ $$$$\mathrm{5}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{of}}\:\mathrm{7}\:\boldsymbol{\mathrm{multiple}}\:\boldsymbol{\mathrm{choice}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{with}} \\ $$$$\mathrm{5}\:\boldsymbol{\mathrm{choices}}? \\ $$ Answered by mr W last updated on 27/Nov/20 $${C}_{\mathrm{5}}…
Question Number 58049 by Aditya789 last updated on 17/Apr/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 123577 by bramlexs22 last updated on 26/Nov/20 Answered by Olaf last updated on 26/Nov/20 $$\begin{cases}{\frac{{g}}{{r}+{g}}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\:\left(\mathrm{1}\right)}\\{\frac{{g}+\mathrm{2}}{\left({r}+\mathrm{4}\right)+\left({g}+\mathrm{2}\right)}\:=\:\frac{\mathrm{10}}{\mathrm{23}}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\::\:{r}+{g}\:=\:\frac{\mathrm{9}}{\mathrm{4}}{g}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\::\:\frac{{g}+\mathrm{2}}{\frac{\mathrm{9}}{\mathrm{4}}{g}+\mathrm{6}}\:\:=\:\frac{\mathrm{10}}{\mathrm{23}} \\ $$$${g}+\mathrm{2}\:=\:\frac{\mathrm{10}}{\mathrm{23}}\left(\frac{\mathrm{9}}{\mathrm{4}}{g}+\mathrm{6}\right) \\ $$$${g}\left(\mathrm{1}−\frac{\mathrm{45}}{\mathrm{46}}\right)\:=\:\frac{\mathrm{60}}{\mathrm{23}}−\mathrm{2}\:=\:\frac{\mathrm{14}}{\mathrm{23}}…
Question Number 57985 by mr W last updated on 15/Apr/19 $${Find}\:{the}\:{number}\:{of}\:{integer}\:{solutions} \\ $$$${for}\:{a}×{b}×{c}×{d}=\mathrm{18900} \\ $$$${with}\:{a},{b},{c},{d}\geqslant\mathrm{1}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19 $$\mathrm{18900}=\mathrm{3}×\mathrm{3}×\mathrm{3}×\mathrm{7}×\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5}…
Question Number 189049 by mr W last updated on 11/Mar/23 Commented by mr W last updated on 13/Mar/23 $${this}\:{is}\:{an}\:{old}\:{question}\:\left({Q}\mathrm{164560}\right).\: \\ $$$${maybe}\:{meanwhile}\:{somebody}\:{has}\:{got}\: \\ $$$${a}\:{new}\:{solution}. \\ $$…
Question Number 189021 by mr W last updated on 10/Mar/23 $${How}\:{many}\:{non}−{similar}\:{triangles} \\ $$$${have}\:{integer}\:{angles}\:{in}\:°? \\ $$ Commented by nikif99 last updated on 11/Mar/23 $${Now}\:{I}\:{think}\:{there}\:{are}\:\mathrm{2700}\:{solutions}\: \\ $$$${for}\:\measuredangle{A},\:\measuredangle{B},\:\measuredangle{C}\:{integers}\:{degrees}.…
Question Number 57909 by mr W last updated on 14/Apr/19 $${n}\:{men}\:{and}\:{n}\:{women}\:{should}\:{be}\:{arranged} \\ $$$${alternately}\:{in}\:{a}\:{row},\:{how}\:{many}\:{ways} \\ $$$${can}\:{this}\:{be}\:{done}?\:{if}\:{the}\:{same}\:{should} \\ $$$${be}\:{done}\:{on}\:{a}\:{table},\:{how}\:{many}\:{ways}\:{then}? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 57688 by Tawa1 last updated on 10/Apr/19 $$\:\:\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{n}:\:\:\:\:\:\:\:\:\underset{\mathrm{i}} {\overset{\mathrm{n}\:−\:\mathrm{1}} {\sum}}\:\:\:\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{i}} \:\mathrm{2}^{\mathrm{i}} \:\:=\:\:\mathrm{65},\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}\:\in\:\mathbb{Z}^{+} .\:\:\:\:\mathrm{where}\:\:\mathrm{zero}\:\mathrm{is}\: \\ $$$$\:\:\mathrm{included} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated…