Hint

i am getting answer as $249$.

Yeah, 249. Simple casework.

Can someone provide a proper solution? I am missing 2 or 3 cases I guess. I am getting 240

I have a solution which I think is probably more faster and easier than the casework thing (this is just my opinion, don't take it personally). So here goes my solution to the problem: The total number of 3-digit numbers in base 10 is just 9*10*10 = 900. Now, the number of 3-digit number in base 10 which do not contain the digit 5 is 8*9*9 = 648. Thus, the number of 3-digit numbers in base 10 having at least one 5 is 900 - 648 = 252. Now, the number of 3-digit numbers in base 10 having at least one 5 and at least two 2's is 3, since any such number would have digits 5, 3, 3 in some order and the number of permutations of 5, 3, 3 is 3!/2! = 3 (note that all the digits 5, 3, 3 are non-zero). Hence, the number of 3-digit numbers in base 10 having at least one 5 and at most one 3 is 252 - 3 = 249