There are total of 5 digits, of which we arrange 3 of them.

Therefore using permutations we can find total possible 3-number combinations.

5P3 = 5*4*3 = 60

Therefore using permutations we can find total possible 3-number combinations.

5P3 = 5*4*3 = 60

To be greater than 600 means first digit must be 6. (1 possibility)

A number is even only if last digit is even, 2 or 4. (2 possibilities)

The 2nd digit then can only be 3,5, (2 or 4). (3 possibilities)

The total possible even 3-digit numbers greater than 600 is product of each digit’s possibilities.

—> 1*3*2 = 6

Therefore probability is 6/60 = 1/10.