A password is created by using $4$ different letters and $4$ different digits in the free arrangements. Note that the password is set in exactly $8$ characters. Two examples for the password are: abcd1234 and 12abcd34 (a and A is considered as same). How many passwords can be created?
A. $C_4^{26} \times C_4^{10}$
B. $C_4^{26} \times C_4^{10} \times 4! \times 4!$
C. $C_4^{26} \times C_4^{10} \times 8!$
D. $(C_4^{26} + C_4^{10}) \times 4!$
E. $(C_4^{26} + C_4^{10}) \times 8!$
I found some opinions, answering the combinatorial problem above. One is the following.
Opinion: See that there are $8$ spaces, $4$ must be filled with distinct letters- and the rest $4$ with distinct digits and all occur simultaneously. So for selecting $4$ distinct letters, no of ways is $26C4$ and setting $4$ distinct digits is $10C4$. Now, you have $8$ distinct objects, so you can permutate them in $8P8=8!$ ways. Thus, the total numbers of password that can be created are: $26C4 \times 10C4 \times 8!$
Is it true? Or maybe you have more powerful arguments to deal with this problem. Please share your idea. Thank you in advance.
