To find the required probability, we need to find the number of ways in which C comes right before the E, and then divide that by the total number of sequences.
For the first, we have:
C E _ _ _ _
_ C E _ _ _
_ _ C E _ _
_ _ _ C E _
_ _ _ _ C E
For each of these cases, we can change the order of the other four letters. For example, for the first one, we can write:
C E A T I V
C E A T V I
C E A I T V
...
We can do that, for each of those 5 cases, a number given by:
4! = 4 * 3 * 2 * 1 = 24
Then, there are 5 * 24 = 120 ways to write C right before the E.
Now, the total number of sequences is given by:
6 * 5 * 4 * 3 * 2 * 1 = 720
This happens because we have 6 possible letters to pick first, then there will rest 5 possible letters to pick after that, and so on.
Finally, the probability we are looking for is:
P = 120/720 = 1/6
Therefore, the second option is correct.