Combinatorics problem: how many such case-sensitive passwords are there?

This problem was on a course exam on mine:

A case sensitive password must contain exactly 4 letters of the English alphabet (26 letters) and exactly 2 numeric digits (0-9) in any order of letters and numbers. How many passwords are there?

Here are some of the things I did understand:

• We need to consider a possible pool of 52 letters (uppercase and lowercase)
• We have 10 digits to choose 2 from

However, the problem is that repetition is allowed (per my teacher's feedback). In other words, the letters we select could all be the same letter (e.g., A).

Here is the answer key:

There are 6 choose 2 ways to arrange 4 letters and 2 numbers, and for each of the four letters there are 52 choices and for each of the numbers there are 10 choices

So essentially, they want you to do:

$6 \choose 2$$52^410^2$

This seems wrong to me, especially the following part:

There are 6 choose 2 ways to arrange 4 letters and 2 numbers

Doesn't this neglect the repetition of letters and/or digits?

• As usual, I think I figured this out by writing out my thoughts on the site. It seems $6 \choose 2$ does account for permuting over multisets, since it translates to $\frac{6!}{2!4!}$. Edit: or maybe not? I don't know. – AleksandrH Feb 23 '18 at 18:42

No. If you choose 4 letters (with repetition) the number of variations are $52^4$.
If repetition wouldn´t be allowed the number of variations would be $52\cdot 51\cdot 50\cdot 49$. Similar for the numbers: $10\cdot 9$
Then you have two types of characters, letters ($4$) and numbers ($2$). They can be arranged in $\binom{4+2}4=\binom{4+2}2=\binom{6}2$ ways.