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An-8-digit-number-divisible-by-9-is-to-be-formed-using-digits-from-0-to-9-without-repeating-the-digits-The-number-of-ways-in-which-this-can-be-done-is-a-72-7i-b-18-7i-c-40-7i-d-36-7i-please-I-m




Question Number 52276 by Necxx last updated on 05/Jan/19
An 8 digit number divisible by 9 is  to be formed using digits from  0 to 9 without repeating the digits.  The number of ways in which this  can be done is?  a)72(7i) b)18(7i) c)40(7i) d)36(7i)    please I mean factorial by i.The  part of my screen where the factorial  symbol is isnt functioning.Thanks
$${An}\:\mathrm{8}\:{digit}\:{number}\:{divisible}\:{by}\:\mathrm{9}\:{is} \\ $$$${to}\:{be}\:{formed}\:{using}\:{digits}\:{from} \\ $$$$\mathrm{0}\:{to}\:\mathrm{9}\:{without}\:{repeating}\:{the}\:{digits}. \\ $$$${The}\:{number}\:{of}\:{ways}\:{in}\:{which}\:{this} \\ $$$${can}\:{be}\:{done}\:{is}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{7}\left.\mathrm{2}\left(\mathrm{7}{i}\right)\:{b}\right)\mathrm{18}\left(\mathrm{7}{i}\right)\:{c}\right)\mathrm{40}\left(\mathrm{7}{i}\right)\:{d}\right)\mathrm{36}\left(\mathrm{7}{i}\right) \\ $$$$ \\ $$$${please}\:{I}\:{mean}\:\boldsymbol{{factorial}}\:{by}\:{i}.{The} \\ $$$${part}\:{of}\:{my}\:{screen}\:{where}\:{the}\:{factorial} \\ $$$${symbol}\:{is}\:{isnt}\:{functioning}.{Thanks} \\ $$
Answered by mr W last updated on 06/Jan/19
we have 10 digits, but we need only 8  from them to form a 8−digit number.  that is to say, we must take 2 of the  10 digits away. but which digits shall  we take away? it is requested that  the 8−digit numbers are divisible by  9. we know a number is divisible by 9  only when the sum of its digits is  divisible by 9. since 0+1+2+...+9=45,  it means when we take all 10 digits,  the numbers formed by them are always  divisible by 9, since 45 is divisible by 9.  when we want to remove two digits to  form 8−digit numbers which remain  divisible by 9, the sum of these two  digits must be divisible by 9. there are  only following 5 possibilities:  we take 0 and 9 away,  we take 1 and 8 away,  we take 2 and 7 away,  we take 3 and 6 away,  we take 4 and 5 away.    when we take 0 and 9 away, with the  remaining 8 digits 1 to 8 we can form  totally 8! numbers.    when we take 1 and 8 away, with the  remaining 8 digits (one of them is 0)  we can form totally only 7×7! numbers  since 0 can not be placed in the first  position.  since we have 4 such similar ways  to take two digits away, we can totally  form 4×7×7! numbers.    so the result is 8!+4×7×7!=36×7!    ⇒answer d) is correct.
$${we}\:{have}\:\mathrm{10}\:{digits},\:{but}\:{we}\:{need}\:{only}\:\mathrm{8} \\ $$$${from}\:{them}\:{to}\:{form}\:{a}\:\mathrm{8}−{digit}\:{number}. \\ $$$${that}\:{is}\:{to}\:{say},\:{we}\:{must}\:{take}\:\mathrm{2}\:{of}\:{the} \\ $$$$\mathrm{10}\:{digits}\:{away}.\:{but}\:{which}\:{digits}\:{shall} \\ $$$${we}\:{take}\:{away}?\:{it}\:{is}\:{requested}\:{that} \\ $$$${the}\:\mathrm{8}−{digit}\:{numbers}\:{are}\:{divisible}\:{by} \\ $$$$\mathrm{9}.\:{we}\:{know}\:{a}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{9} \\ $$$${only}\:{when}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is} \\ $$$${divisible}\:{by}\:\mathrm{9}.\:{since}\:\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{9}=\mathrm{45}, \\ $$$${it}\:{means}\:{when}\:{we}\:{take}\:{all}\:\mathrm{10}\:{digits}, \\ $$$${the}\:{numbers}\:{formed}\:{by}\:{them}\:{are}\:{always} \\ $$$${divisible}\:{by}\:\mathrm{9},\:{since}\:\mathrm{45}\:{is}\:{divisible}\:{by}\:\mathrm{9}. \\ $$$${when}\:{we}\:{want}\:{to}\:{remove}\:{two}\:{digits}\:{to} \\ $$$${form}\:\mathrm{8}−{digit}\:{numbers}\:{which}\:{remain} \\ $$$${divisible}\:{by}\:\mathrm{9},\:{the}\:{sum}\:{of}\:{these}\:{two} \\ $$$${digits}\:{must}\:{be}\:{divisible}\:{by}\:\mathrm{9}.\:{there}\:{are} \\ $$$${only}\:{following}\:\mathrm{5}\:{possibilities}: \\ $$$${we}\:{take}\:\mathrm{0}\:{and}\:\mathrm{9}\:{away}, \\ $$$${we}\:{take}\:\mathrm{1}\:{and}\:\mathrm{8}\:{away}, \\ $$$${we}\:{take}\:\mathrm{2}\:{and}\:\mathrm{7}\:{away}, \\ $$$${we}\:{take}\:\mathrm{3}\:{and}\:\mathrm{6}\:{away}, \\ $$$${we}\:{take}\:\mathrm{4}\:{and}\:\mathrm{5}\:{away}. \\ $$$$ \\ $$$${when}\:{we}\:{take}\:\mathrm{0}\:{and}\:\mathrm{9}\:{away},\:{with}\:{the} \\ $$$${remaining}\:\mathrm{8}\:{digits}\:\mathrm{1}\:{to}\:\mathrm{8}\:{we}\:{can}\:{form} \\ $$$${totally}\:\mathrm{8}!\:{numbers}. \\ $$$$ \\ $$$${when}\:{we}\:{take}\:\mathrm{1}\:{and}\:\mathrm{8}\:{away},\:{with}\:{the} \\ $$$${remaining}\:\mathrm{8}\:{digits}\:\left({one}\:{of}\:{them}\:{is}\:\mathrm{0}\right) \\ $$$${we}\:{can}\:{form}\:{totally}\:{only}\:\mathrm{7}×\mathrm{7}!\:{numbers} \\ $$$${since}\:\mathrm{0}\:{can}\:{not}\:{be}\:{placed}\:{in}\:{the}\:{first} \\ $$$${position}. \\ $$$${since}\:{we}\:{have}\:\mathrm{4}\:{such}\:{similar}\:{ways} \\ $$$${to}\:{take}\:{two}\:{digits}\:{away},\:{we}\:{can}\:{totally} \\ $$$${form}\:\mathrm{4}×\mathrm{7}×\mathrm{7}!\:{numbers}. \\ $$$$ \\ $$$${so}\:{the}\:{result}\:{is}\:\mathrm{8}!+\mathrm{4}×\mathrm{7}×\mathrm{7}!=\mathrm{36}×\mathrm{7}! \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{d}\right)\:{is}\:{correct}. \\ $$
Commented by Necxx last updated on 06/Jan/19
wow.....Thank you sir.
$${wow}…..{Thank}\:{you}\:{sir}. \\ $$

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