The digits 2 comma 3 comma 4 comma 5 comma and 2, 3, 4, 5, and 6 are randomly arranged to form a three-digit number. (Digits are not repeated.) Find the probability that the number is even and greater than 600.

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

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.