Password combinatorics on identical notes

Question

I have the following two questions:

A password consists only small letters a,…,z

1. How many possibilities are there to put in a hat two identical folded notes such that on one of them is a password of length 6 and on the other a password of length 7, such that no letter repeats twice (not on the same note, and not between the two notes)? Is the answer different when the notes are of different colors? Explain.

2. How many possibilities are there to put in a hat two identical folded notes such that on each of them is a password of length 6? Is the answer different when the notes are of different colors?

My answers are as follows:

1. The notes are identical, hence we can generate a length 13 password and "split" it in the middle to 6 and 7 length password with no repeats, hence the answer is $\frac{26!}{13!}$. When the notes are not identical, we need to multiply the result by 2 as once one colored note has the 6 length pass, and the over the 7 and vice versa. Hence $2(\frac{26!}{13!})$
2. The letters can repeat in the password itself and on the other note, but the notes are identical hence $(passA, passB) and (passB, passA) \mapsto \left \{ passA, passB \right \}$. Meaning we map 2 to 1, and the answer is $\frac{26^{12}}{2}$. When the colors are different we map 1 to 1 hence the answer is $26^{12}$.

We had a debate between us students whether the above are correct, some say the calculations are wrong, some say the colors don't make any difference in the first problem...

Any help would be greatly appreciated!

Answer 1

How many possibilities are there to put in a hat $2$ notes such that on one of them is a password of length $6$ and on the other a password of length $7$, such that no letter repeats more than once?

If the notes are identical: $\binom{26}{6+7}\cdot(6+7)!$

If the notes are different: $\binom{26}{6+7}\cdot(6+7)!\cdot2!$

How many possibilities are there to put in a hat $2$ notes such that on each one of them is a password of length $6$?

If the notes are identical: $26^{6+6}/2!$

If the notes are different: $26^{6+6}$

Answer 2

Case 1-

We have to pick 13 alphabets total options 26.

Ways = $P(26,7) \times P(19,6) + P(26,6) \times P(20,7)$

In second case -

Number of ways = $P(26,6) \times P(20,6) + P(26,6) \times P(20,6)$

As we have two colours.

Both answers are not same as in one we are picking 13 alphabets and in other 12 alphabets.