How many 6 character passwords are possible if letters cannot be repeated but digits CAN be? A computer system uses passwords that have six characters, and each character is
one of the 26 letters (a-z) or 10 integers (0-9). The first character has to be a letter.
The passwords are not case sensitive. How many contain no repeated letters but can contain repeated integers?
I could do this question easily if neither letters nor integers could be repeated but having one allow repetition and the other not has thrown me off.
Any help would be greatly appreciated.
Thanks
 A: Edited to Joffan's input
How I would guess to approach this, I am also pretty new to combinatorics.
For the first space you have 26 options since there is NOT case sensitive.
(26 option)(..)(..)(..)(..)(..)
Now for the remaining five slots you would have six possible cases.
$$ \begin{array}{|c|c|} \hline
\text{# Letters (after first)} & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline
\text{# Numbers} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
\end{array}$$
There is another layer to think about say in the case of three letter and two numbers, you know have five slots and three slots to place letters so you have ${5 \choose 3}$ ways of picking positions for letters and for each of the positions you would have 25 options for the first slots since one letter is taken for the fist slot, 24 options for the second and 23 options for the third slots, as for the last two number slots you would have $10\cdot 10$ options.
I would then take a guess that the final number would be
$26 \cdot \left({5 \choose 5} 
\cdot \frac{25!}{(25-5)!} \cdot 10^0 + {5 \choose 4} 
\cdot\frac{25!}{(25-4)!} \cdot 10^1 + {5 \choose 3} 
\cdot\frac{25!}{(25-3)!} \cdot 10^2 +{5 \choose 2} 
\cdot\frac{25!}{(25-2)!} \cdot 10^3 +{5 \choose 1} 
\cdot\frac{25!}{(25-1)!} \cdot 10^4 +{5 \choose 0} 
\cdot\frac{25!}{(25)!} \cdot 10^5\right)$
or rather
$26 \cdot \sum_{k=0}^{5}\left({{5 \choose k} 
\cdot\frac{25!}{(25-k)!}} \cdot 10^{5-k}\right)$.
And as a final summary since the multiplicative factor of 26 distributes to all summands we end up with
$$\sum_{k=0}^{5}\left({{5 \choose k} 
\cdot\frac{26!}{(25-k)!}} \cdot 10^{5-k}\right)$$
This is my first time posting so I know my answer is poorly formatted, I cannot fully verify my answer but this is what I would turn in if I had to give it an attempt. I would love to get some feedback on this as I recently started self learning combinatorics.
