Question

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.

Answer 1

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.