where 



Hence the values of betwwen 0 and 360 are [when n=1,n=2]
First we will find the number of three digit numbers (i.e, numbers from 100 to 999)which can be formed using the digits 0,1,2,3,4,5,6,7,8and 9 with repitition allowed .
Now we have the first place can be filled by any of the 9 digits other than 0 and since repetition is allowed the second and third can be filled by any of the ten digits.
Hence the total number of three digit numbers will be =
Now we will consider the case that the number does not have the digit 5.
Now the first place can be filled by any of the 8 digits other than 0 and 5 and since repetition is allowed the second and third can be filled by any of the 9 digits other than 5.
Hence the total number of ways we can form a three digit number with out 5 will be =
Therefore the number of three digit numbers with atleast one 5=