If a 3-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number (which is a 3 -digit number) is divisible by 4 and 5 ?

We need 3 -digit number which is divisible by $4 \& 5$ both. i.e. their last two digits are $00,20,40,60 \& 80$ Now, ending with 00 are '9'. $\{100,200, \ldots \ldots, 900\}$ If digit repeat other than ' 0 ' then they are $\{220,440,660,880\}$ but 220 numbers can be permuted according to the condition as $\{220,202\}$ So, there are ' 8 ' other favorable cases. If the number have no digit repeated like 320 . 320 can be permuted in 4 ways. $\{302,230,320,203\}$ So, such numbers are $8 \times 4 \times 4=128$ Total favorable $=9+8+128=145$ So, required prob. $=\frac{145}{900}=\frac{29}{180}$