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