Question

An eight 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. $40(7!)$
• B. $36(7!)$
• C. $18(7!)$
• D. $72(7!)$

Answer 1

The correct option is B $36(7!)$

$S=\{0,1,2,3,4,5,6,7,8,9\}\to 10digits$

${\sum }_0^{9}i=45\Rightarrow$ divisible by 9

$\therefore$ For $8$ digit number we need to remove two digits from $S$

After removing $\sum \Rightarrow$ divisible by 9

$\therefore$ We can only remove the pairs $(0,9),(1,8),(2,7),(3,6),(4,5)$

Since $0+9=9,45-9=36\Rightarrow$ divisible by 9

$1+8=9,45-9=36\Rightarrow$ divisible by 9

$4+5=9$

$3+6=9$

$\therefore$ If $(0,9)$ are removed then no. of 8 digits nos possible $=8!$

if $(1,8)$ are removed the no. of 8 digit nos $=8!-7!$ (subtracting the number of cases where '0' is at the left most place)

Similarly, when we remove $(2,7)$, $(3,6)$ and $(4,5)$ we get $8!-7!$ in each case.

$\therefore$ Total 8 digit nos $=8!+4(8!-7!)=5\cdot 8!-4\cdot 7!$

$=40\cdot 7!-4\cdot 7!$

$=36(7!)$