How many possible passwords of 6 characters (only lower-case letters and digits) can be made with only 2 lower-case letters and no repeated digits? 
How many possible passwords of 6 characters (only lower-case letters and digits) can be made with only 2 lower-case letters and no repeated digits?

My solution for this is
$${}_6C_2 \cdot 26 \cdot 26 \cdot 10 \cdot 9 \cdot 8 \cdot 7 = 51105600$$
However, my lecturer states that it is
$$26 \cdot 26 \cdot {}_{10}C_4 \cdot 6!= 102211200$$
Can someone explain why this is the case? Or is there another solution?
 A: I believe you are correct, and that your lecturer's solution overcounts the solution by a factor of two exactly.
By looking at the formulas provided, it seems that your reasoning is as follows: choose two of the six spaces to be letters, choose the letters used, then choose the numbers used in the remaining four spaces. Hence, $\binom{6}{2} \cdot 26^2 \cdot 10 \cdot 9 \cdot 8 \cdot 7$.
On the other hand, your lecturer's reasoning seems to be: choose which letters and numbers to use first, then permute the characters arbitrarily. Hence, $26^2 \cdot \binom{10}{4} \cdot 6!$. However, the problem with this approach is that it overcounts the number of distinct ways you can permute the letters. For example, if two $a$'s were chosen for the letters (denote then $a_1$ and $a_2$ to distinguish them from each other) and, say, $1234$ were chosen for the digits, the passwords $$a_1a_21234 \quad \text{and} \quad a_2a_11234$$ would be counted differently, even though they're the same password. This doesn't happen in your case, since fixing the positions of the letters first before choosing them results in each pair of letters leading to a different password combination.
A: Congratulations !
You are correct, and your lecturer is wrong !
For the lecturer approach to be corrected, it would need to be
$\binom{10}4*[6!*\left(\binom{26}2 +26/2\right) ]$
but why think of such a complex formulation !
A: CASE 1. One letter repeats twice. Select the letter in 26 ways. Choose 4 digits in $\binom{10}{4}$ ways. Arrange them (assuming 0 can take the first position since it is just a password) in $\frac{6!}{2!}$ ways. So here we have $26. \binom{10}{4} \frac{6!}{2!}= 1965600$.
CASE 2. Two distinct letters are employed. Select the letters in $\binom{26}{2}$ ways. Choose 4 digits in $\binom{10}{4}$ ways. Arrange them (assuming 0 can take the first position since it is just a password) in 6! (No repetition)ways. Total = $\binom{26}{2}\binom{10}{4}6! = 49140000$
So total is $1965600+49140000=51105600. $
