Solution:
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(7!) ways due to presence of 0.
So, total number of ways =8!+4×7(7!)=36(7!)
© examsnet.com