Question Number 52276 by Necxx last updated on 05/Jan/19
$${An}\:\mathrm{8}\:{digit}\:{number}\:{divisible}\:{by}\:\mathrm{9}\:{is} \\ $$$${to}\:{be}\:{formed}\:{using}\:{digits}\:{from} \\ $$$$\mathrm{0}\:{to}\:\mathrm{9}\:{without}\:{repeating}\:{the}\:{digits}. \\ $$$${The}\:{number}\:{of}\:{ways}\:{in}\:{which}\:{this} \\ $$$${can}\:{be}\:{done}\:{is}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{7}\left.\mathrm{2}\left(\mathrm{7}{i}\right)\:{b}\right)\mathrm{18}\left(\mathrm{7}{i}\right)\:{c}\right)\mathrm{40}\left(\mathrm{7}{i}\right)\:{d}\right)\mathrm{36}\left(\mathrm{7}{i}\right) \\ $$$$ \\ $$$${please}\:{I}\:{mean}\:\boldsymbol{{factorial}}\:{by}\:{i}.{The} \\ $$$${part}\:{of}\:{my}\:{screen}\:{where}\:{the}\:{factorial} \\ $$$${symbol}\:{is}\:{isnt}\:{functioning}.{Thanks} \\ $$
Answered by mr W last updated on 06/Jan/19
$${we}\:{have}\:\mathrm{10}\:{digits},\:{but}\:{we}\:{need}\:{only}\:\mathrm{8} \\ $$$${from}\:{them}\:{to}\:{form}\:{a}\:\mathrm{8}−{digit}\:{number}. \\ $$$${that}\:{is}\:{to}\:{say},\:{we}\:{must}\:{take}\:\mathrm{2}\:{of}\:{the} \\ $$$$\mathrm{10}\:{digits}\:{away}.\:{but}\:{which}\:{digits}\:{shall} \\ $$$${we}\:{take}\:{away}?\:{it}\:{is}\:{requested}\:{that} \\ $$$${the}\:\mathrm{8}−{digit}\:{numbers}\:{are}\:{divisible}\:{by} \\ $$$$\mathrm{9}.\:{we}\:{know}\:{a}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{9} \\ $$$${only}\:{when}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is} \\ $$$${divisible}\:{by}\:\mathrm{9}.\:{since}\:\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{9}=\mathrm{45}, \\ $$$${it}\:{means}\:{when}\:{we}\:{take}\:{all}\:\mathrm{10}\:{digits}, \\ $$$${the}\:{numbers}\:{formed}\:{by}\:{them}\:{are}\:{always} \\ $$$${divisible}\:{by}\:\mathrm{9},\:{since}\:\mathrm{45}\:{is}\:{divisible}\:{by}\:\mathrm{9}. \\ $$$${when}\:{we}\:{want}\:{to}\:{remove}\:{two}\:{digits}\:{to} \\ $$$${form}\:\mathrm{8}−{digit}\:{numbers}\:{which}\:{remain} \\ $$$${divisible}\:{by}\:\mathrm{9},\:{the}\:{sum}\:{of}\:{these}\:{two} \\ $$$${digits}\:{must}\:{be}\:{divisible}\:{by}\:\mathrm{9}.\:{there}\:{are} \\ $$$${only}\:{following}\:\mathrm{5}\:{possibilities}: \\ $$$${we}\:{take}\:\mathrm{0}\:{and}\:\mathrm{9}\:{away}, \\ $$$${we}\:{take}\:\mathrm{1}\:{and}\:\mathrm{8}\:{away}, \\ $$$${we}\:{take}\:\mathrm{2}\:{and}\:\mathrm{7}\:{away}, \\ $$$${we}\:{take}\:\mathrm{3}\:{and}\:\mathrm{6}\:{away}, \\ $$$${we}\:{take}\:\mathrm{4}\:{and}\:\mathrm{5}\:{away}. \\ $$$$ \\ $$$${when}\:{we}\:{take}\:\mathrm{0}\:{and}\:\mathrm{9}\:{away},\:{with}\:{the} \\ $$$${remaining}\:\mathrm{8}\:{digits}\:\mathrm{1}\:{to}\:\mathrm{8}\:{we}\:{can}\:{form} \\ $$$${totally}\:\mathrm{8}!\:{numbers}. \\ $$$$ \\ $$$${when}\:{we}\:{take}\:\mathrm{1}\:{and}\:\mathrm{8}\:{away},\:{with}\:{the} \\ $$$${remaining}\:\mathrm{8}\:{digits}\:\left({one}\:{of}\:{them}\:{is}\:\mathrm{0}\right) \\ $$$${we}\:{can}\:{form}\:{totally}\:{only}\:\mathrm{7}×\mathrm{7}!\:{numbers} \\ $$$${since}\:\mathrm{0}\:{can}\:{not}\:{be}\:{placed}\:{in}\:{the}\:{first} \\ $$$${position}. \\ $$$${since}\:{we}\:{have}\:\mathrm{4}\:{such}\:{similar}\:{ways} \\ $$$${to}\:{take}\:{two}\:{digits}\:{away},\:{we}\:{can}\:{totally} \\ $$$${form}\:\mathrm{4}×\mathrm{7}×\mathrm{7}!\:{numbers}. \\ $$$$ \\ $$$${so}\:{the}\:{result}\:{is}\:\mathrm{8}!+\mathrm{4}×\mathrm{7}×\mathrm{7}!=\mathrm{36}×\mathrm{7}! \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{d}\right)\:{is}\:{correct}. \\ $$
Commented by Necxx last updated on 06/Jan/19
$${wow}…..{Thank}\:{you}\:{sir}. \\ $$