We need a 3-digit number which is divisible by 4 and 5 both.
i.e. Their last two digits are: 00, 20, 40, 60 and 80.
Now, ending with 00 are:
{100, 200, 900}
If a digit repeats other than '0', then they are:
{220, 440, 660, 880}
But 220 can be permuted according to the condition as {220, 202}.
So, there are '8' other favorable cases.
If the number has no digit repeated, like 320:
320 can be permuted in 4 ways.
{302, 230, 320, 203}
So, such numbers are: 8 x 4 x 4 = 128
Total favourable cases = 9 + 8 + 128 = 145
So, required probability = 145/900 = 29/180