I'd say 100 even, don't know if this is a real question or not.
The conclusion is valid
it is a systematic random sample
and the second one is a line graph (I think)
If the letters are considered distinct, then the number of permutations is

.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or

.
A =
5 and -5
-1 and 3
¹/₃B =
-6 and 12
-7 and 7
¹/₃B + A =
-1 and 7
-8 and 10