There is a password that contains exactly 6 characters consisting of: 3 Digits, 2 Letters and 1 Special characters
Assuming there is only 7 special characters, how many different ordered arrangements of these characters are there (If password is case sensitive and Repetition is not allowed)
Is my solution correct?: 10 . 9 . 8 . 52 . 51 . 7 = 13 66 080 combinations
Or:
10p3 . 52p2 . 7p1 = 13 366 080 combinations
Neither.
Your solution(s) (it is really the same solution; the first number misses a $3$) would work if you have to start with the 3 digits, followed by the 2 letters, and ends with the special character, but presumably you can mix those up in any order.
So, instead I would pick the 3 digits using the choose function (i.e. ${10 \choose 3}$), pick the 2 letters and the special symbol likewise, and then multiply by the number of permutations of those $6$ characters, i.e. $6!$. Thus, you get:
$${10 \choose 3} \cdot {52 \choose 2} \cdot {7 \choose 1} \cdot 6!$$