Choosing code using digits and letters 
Want to put together a secret code. The code consists of 2 different digits and 3 different English letters (26 options).
  How many different codes can be put together?

I tried to think of it this way:
$$10*9*26*25*24=1,404,000$$
Because I first choose two digits and then three letters.
The answer by the book is:
$$\binom{10}{2}\binom{26}{3}*5!=14,040,000 $$
I understand why they did it, but I do not know what I'm missing to get answer similar to their own.
Thank you very much.
 A: Before you can put the code together, you need to choose which spaces are going to be numbers and which spaces are going to be letters. There are $5$ spaces altogether (since you are making a code from $5$ symbols) and there are $2$ spaces for numbers. Therefore, there are $5 \choose 2$ ways to pick which spaces will be where the numbers go and which spaces will be where the letters go. Thus, your answer should be:
$${5 \choose 2}*10*9*26*25*24=14,040,000$$
A: Algebraically,
$$
\binom{10}{2}\binom{26}{3}5! = \frac{10!}{2!8!}\frac{26!}{23!3!} = \frac{10\times 9 \times 26 \times 25 \times 24 \times 5!}{2!3!} =  10\times 9 \times 26 \times 25 \times 24 \times 10 .
$$
Te difference is that although you considered all the possible choices of letters and digits, you forgot to consider all the possible ways to arrange the chosen items, which is $5!$. 
A: Hint. You counted the codes in which the digits come first, followed by the numbers.
A: Obviously something's fishy with the first slot, as it can't be 0. Let's split the solution into 2: 
1) First slot is a letter. We have 26 options for the first slot, then choose two more for letters: $\binom{4}{2} \times 25 \times 24$, then multiply by $10 \times 9$.
2) First slot is a digit: We have 9 options for it, then we choose 4 slots for the second digit: $9 \times \binom{4}{1} \times 9$ and the rest are chars. The full solution is 
$$
26 \times \binom{4}{2} \times 25 \times 24 \times 10 \times 9 +9 \times \binom{4}{1} \times 9 \times 26 \times 25 \times 24
$$ 
