# 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).

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.