Question Number 109387 by mr W last updated on 19/Sep/20 $$\boldsymbol{\mathrm{SUCCESSFULLY}} \\ $$$${How}\:{many}\:{different}\:{words}\:{can}\:{you} \\ $$$${form}\:{using}\:{these}\:{letters}\:{so}\:{that}\:{no} \\ $$$${two}\:{same}\:{letters}\:{are}\:{adjacent}? \\ $$ Commented by bobhans last updated on…
Question Number 43824 by Necxx last updated on 15/Sep/18 Answered by MJS last updated on 16/Sep/18 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\sqrt{\pi} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}!=\mathrm{1} \\ $$$$\left(\mathrm{B}\right)\:\:\mathrm{sin}\:\mathrm{45}°\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left(\mathrm{C}\right)\:\:\sqrt{\mathrm{arccos}\:−\mathrm{1}}=\sqrt{\pi}…
Question Number 174855 by byaw last updated on 12/Aug/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\ $$$$\mathrm{a}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{4}\:\mathrm{people}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{5}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{7}\:\mathrm{women}\:\mathrm{when}\:\mathrm{it}\:\mathrm{contains} \\ $$$$\mathrm{people}\:\mathrm{of}\:\mathrm{both}\:\mathrm{sexes}\:\mathrm{and} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{as}\:\mathrm{many} \\ $$$$\mathrm{women}\:\mathrm{as}\:\mathrm{men}. \\ $$ Answered…
Question Number 43659 by Joel578 last updated on 13/Sep/18 $$\mathrm{An}\:\mathrm{unfair}\:\mathrm{coin}\:\mathrm{with}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{getting} \\ $$$$\mathrm{head}\:\mathrm{in}\:\mathrm{one}\:\mathrm{toss}\:=\:\frac{\mathrm{1}}{\mathrm{5}}. \\ $$$$\mathrm{If}\:\mathrm{coin}\:\mathrm{tosses}\:{n}\:\mathrm{times},\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of} \\ $$$$\mathrm{getting}\:\mathrm{2}\:\mathrm{heads}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\: \\ $$$$\mathrm{getting}\:\mathrm{3}\:\mathrm{heads}.\:\mathrm{Find}\:{n} \\ $$ Commented by math1967 last updated…
Question Number 43343 by pieroo last updated on 10/Sep/18 $$\mathrm{how}\:\mathrm{many}\:\mathrm{odd}\:\mathrm{numbers}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{60000}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9},\mathrm{0}\:\mathrm{if}\:\mathrm{no}\:\mathrm{number}\:\mathrm{contains} \\ $$$$\mathrm{any}\:\mathrm{digit}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once}? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18 $$\underset{\mathrm{5}} {−}\:\underset{\mathrm{4}}…
Question Number 108638 by bemath last updated on 18/Aug/20 $$\:\:\:\frac{{bemath}}{\bigstar} \\ $$$${prove}\:{that}\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix} \\ $$ Answered by bemath last updated on 18/Aug/20 Terms of Service Privacy…
Question Number 43031 by aseerimad last updated on 06/Sep/18 Commented by MrW3 last updated on 06/Sep/18 $$\mathrm{8}!×\left(\mathrm{9}+{C}_{\mathrm{2}} ^{\mathrm{9}} \right)=\mathrm{8}!×\mathrm{45}=\mathrm{5}×\mathrm{9}!=\mathrm{1814400} \\ $$ Commented by MrW3 last…
Question Number 108450 by bemath last updated on 17/Aug/20 $$\:\:\frac{\mathcal{B}{e}\mathcal{M}{ath}}{\subset\supset} \\ $$$$\left(\mathrm{1}\right){find}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)! \\ $$$$\left(\mathrm{2}\right)\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{x}\:\mathrm{sin}\:{x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$ Commented by Smail last updated on…
Question Number 173976 by savitar last updated on 22/Jul/22 $$ \\ $$$$\:\:\:\:{B}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\Gamma\left({s}\right)=\:\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$ \\…
Question Number 173975 by savitar last updated on 22/Jul/22 $$ \\ $$$$\: \\ $$$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{y}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{y}\right)^{{b}} }\:{dy}\:\overset{{u}=\frac{\mathrm{1}}{\mathrm{1}+{y}}} {=}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{u}\right)^{{a}−\mathrm{1}} }{{u}^{{a}−\mathrm{1}} }\:{u}^{{b}}…