Ankit Jaiswal
Last Activity: 8 Years ago
Hi,
I hope this helps.
As the number should be divisible by 4 it must end with 12,16,32,36,52,56 and hence there are a total of 6 cases.
Now in all the cases there are remaining 2 digits and and 3 different numbers so the number of ways to arrange them = 3!/(3-2)! = 3! = 3*2 = 6
and hence for all cases total numbers possible will be 6*6 (6 cases and 6 numbers in each case)
=
Thanks
