discrete mathematics notes possibilities problem 
In how many ways you can put two similar notes in a bag, so on each
  one there is a password  of 6 letters (abc..z).

I am having trouble with it. My way of thinking:
On the first note: 26^6
On the second note: it is the same thing, but i may have doubles, so i need to divide it by 2. so i am thinking of (26^6)/2
so the answer is 26^6 + (26^6)/2
But i don't think I am right.
 A: For the first bag: There are $6$ possible spaces and a total of $26$ possible letters. So, the number of ways to create a password is $26^6$.
For the second bag: There are again $6$ spaces and $26$ letters, so the number of ways are $26^6$.
The total number of ways are thus $26^6 \times 26^6 = 26^{12}$.
But, in case you want the passwords in the two bags to be different then, the second bag should not contain the password as that of the first one, so we should reduce it by $1$, that is, the total ways only under this condition for the second bag are $26^6-1$. Hope it helps.
A: You need to specify whether or not the notes must list 6 distinct letters.
You also need to define "similar" when comparing two passwords. Is differing by only one letter sufficient? 
Assuming that letters can appear more than once in a given password, which I assume to be the case, given your answer about the first note), then your answer regarding the the first note is correct. 
But your answer about the second bag should be $26^6 - 1$,  where we must subtract the one password that is identical to the password in the first note.  
So:  In how many ways you can put two non-identical notes in a bag, so on each one there is a password of 6 letters (abc..z)?

$$26^6(26^6 -1)$$

A: There are $((26)^6)^2=26^{12}$ such configurations possible, if no restrictions for certain passwords apply.
