NIMCET 2008 Question Paper with solutions

Sum of the digits of a number divisible by 9 is also divisible by 9 .
Now, sum of digits $0,1,2,3,...,9$ is 45 which is divisible by 9
So, two digits out of 10 digits given will be omitted in such a way that their sum should also be divisible by $9.$
So, omitted digits will be (0,9),(1,8),(2,7),(3,6) and (4,5) . In the first case, there will be 8 ! numbers divisible by 9 and in the last four cases thero will be $7\left(7!\right)$ ways due to presence of $0.$
So, total number of ways $=8!+4\times 7\left(7!\right)=36\left(7!\right)$