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An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5,6,7,8,9 without replacement.
The number of ways in which can be done is 


Options:
A .   2(7!)
B .   4(7!)     
C .   (36)(7!)
D .   (7!)(33)
E .   3q+ r
Answer: Option C
:
C

From the digits 0 1 2 3 4 5 6 7 8 9


The sum is 45


Thus, every time we remove 2 digits which add up to 9, we will have a new set of 8 numbers which are divisible by 9


Case 1


Removing 0 and 9


Number of cases possible = 8!


Case 2


Removing


(1)8 and 1


(2)7 and 2


(3)6 and 3


(4)5 and 4


In each case, total number of cases = 8!-7!


(we subtract 7! Because those are the cases where 0 is the first digit)


Hence, total number of ways = 4(8!-7!)+8!= 36x7!. 



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