You're right that this is not the right board, but then, I'm not sure either since one thread similar to this got moved from the simple probability and statistics board, and I do factorials, permutations and combinations in statistics.
To answer your first question:
Here, there are repeats, and as such, is different. My way of solving this is this:the number of ways you can use capital and lowercase a's
aa AA aA Aa
theres only two variations, a and A. 2*1=2.
I first put two blanks: _ _
In the first blank, I can have either a or A, so, two possibilities.
In the second blank, I can get either a or A, so again two possibilities.
Total becomes 2 x 2 = 4
This works with more than two items too.
With abc, you can have:
aaa, aab, aba, aac, aca, abc, acb, abb, acc,
baa, bab, bba, bbb, bac, bca, bcc, bcb, bbc
caa, cab, cba, cbb, cac, cca, ccc, ccb, cbc
Which is obtained using the logic; 3 possibilities in each blank, hence, 3 x 3 x 3 = 27
IF there were no repeats, then 2! is the answer for the first case, and 3! for the second case.
Oh, I see I already answered your next question