How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which one divisible by 5 and none of the digit is repeated?

Since 3 digit number is divisible by 5.

So we must have 5 at the unit place, so then is 1 way the tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9) there are 5 ways filling trans place the hundred place can now be filled by any of the meaning 4 digit.

Required number of numbers $=1\times 5\times 4=20$