License plates with no repetition If there are no restrictions on where the digits and letters are placed. How many $8$-place license plate consisting of $5$ letters and $3$ digits are possible if no repretitions of letters or digits are allowed..?
Naturally I write
$26*25*24*23*22*10*9*8$
This is the count if we want them to be in order. Since there no restictions we can permute them any way we want and that gives  8!. So the answer should be the product of the numbers. However, the answer key says the correct answer is
$$ {26 \choose 5} {10 \choose 3} 8! $$
I believe this to be a typo since repetitions is not allowd
 A: If no repetitions of letters then we have $\dbinom{26}{5}$ ways.
If no repetitions of digits then we have $\dbinom{10}{3}$ ways.
If no repetitions of letters or digits are allowed, then we have $$\dbinom{10}{3}\dbinom{26}{5}\cdot8!\mbox{ ways}$$
A: As Dominik Kutek points out in the comments, the number $26\cdot25\cdot24\cdot23\cdot22\cdot10\cdot9\cdot8$ gives the number of ways to choose 5 distinct letters and 3 distinct digits, where order matters among the letters and among the digits.  E.g. this number counts the choice of ABCDE and 123 as distinct from EDCBA and 321.  Since there are no restrictions on where letters and digits can occur in the string of eight symbols, we must multiply by the number of configurations of 5 letters and 3 digits.  Among the eight positions, we must choose five positions for the letters, with the remaining positions filled in by the digits.  There are ${8\choose 5}$ ways to do this, so the answer is $26\cdot25\cdot24\cdot23\cdot22\cdot10\cdot9\cdot8 \cdot {8\choose 5}$.  Note that this answer agrees with the answer key:
$$(26 \cdot 25 \cdot 24 \cdot 23 \cdot 22) (10 \cdot 9 \cdot 8) {8\choose 5} \\
= \left( \frac{26!}{21!} \right) \left(\frac{10!}{7!}\right) \left(\frac{8!}{5! 3!}\right) \\
= {26 \choose 5}5! {10 \choose 3}3! \left(\frac{8!}{5! 3!}\right) \\
= {26 \choose 5} {10 \choose 3} 8!
$$
A: The answer key is right. The first factor is the number of ways to choose five distinct letters from $26$. The second factor counts digit choices. The factorial counts arrangements.
