Question Number 110523 by Aina Samuel Temidayo last updated on 29/Aug/20 $$ \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{square}\:\mathrm{PQRS},\:\mathrm{K}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{midpoint}\:\mathrm{of}\:\mathrm{PQ},\:\mathrm{L}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{QR},\:\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{RS},\:\mathrm{N}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{midpoint}\:\mathrm{of}\:\mathrm{SP}\:\mathrm{and}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint} \\ $$$$\mathrm{of}\:\mathrm{KM}.\:\mathrm{A}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{from} \\ $$$$\mathrm{each}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{points}\:\mathrm{from} \\…
Question Number 110498 by Aina Samuel Temidayo last updated on 29/Aug/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{word}\:\mathrm{MATHEMATICS}\:\mathrm{be} \\ $$$$\mathrm{rearranged}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{word}\:\mathrm{formed} \\ $$$$\mathrm{either}\:\mathrm{starts}\:\mathrm{or}\:\mathrm{ends}\:\mathrm{with}\:\mathrm{a}\:\mathrm{vowel},\:\mathrm{and} \\ $$$$\mathrm{any}\:\mathrm{three}\:\mathrm{consecutive}\:\mathrm{letters}\:\mathrm{must} \\ $$$$\mathrm{contain}\:\mathrm{a}\:\mathrm{vowel}? \\ $$ Commented…
Question Number 175567 by mr W last updated on 02/Sep/22 $${in}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{put}\:\mathrm{40} \\ $$$${identical}\:{balls}\:{into}\:\mathrm{20}\:{identical}\:{boxes} \\ $$$${such}\:{that}\:{each}\:{box}\:{obtains}\:{at}\:{least}\:{one} \\ $$$${ball}\:{and}\:{at}\:{most}\:\mathrm{5}\:{balls}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 109488 by bubugne last updated on 24/Aug/20 $${How}\:{many}\:{are}\:{the}\:{permutations}\:{of} \\ $$$$\mathrm{1}\:−\:{a}\:{little}\:{rubik}'{s}\:{cube}\:{with}\:\mathrm{4}\:{squares}\:{by}\:{side} \\ $$$$\mathrm{2}\:−\:{a}\:{classical}\:{one}\:{with}\:\mathrm{9}\:{squares}\:{by}\:{side} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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…