How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits? Can somebody please explain why don't we count the arrangements of $7$ (digits and letters) as $7!$? In my understanding, it will be $7!\cdot 26^3 \cdot 10^4$ I know it's wrong and the solution tells us $35$ possible letters arrangements and not $7!$ but I can't understand why.
 A: Let's assume the license plate has the 3 letters on the left hand side in a row, and the four letters on the right hand side in a row, like this:
$$\mathrm{AAA}1234$$
We have have 26 possible letters to choose from, and we make this choice three times. So this gives us $26^3$ possible letter combinations. Next, we have 10 single-digit numbers to choose from, and we make this choice 4 times. This gives us $10^4$ possible number combinations. Multiplying these gives us $26^{3}\times 10^4$ possible plates.
On the other hand, if there is no restriction on the placement of letters and numbers, we could have something like this: $$\mathrm{A1A2A34B}$$ But we can still only have 3 letters and 4 numbers.
So, to start out let's decide where to put our three letters. We need to pick 3 places out of 7, so thats $7 \choose 3$ possibilities. Now that we have decided where to put our three letters, the other four spots must be numbers. At this point we have returned to our original problem where we know the positions of the letters and numbers. So the final answer should be $${7 \choose3} \times 26^3 \times 10^4$$
I'm not sure where the $7!$ comes from in your solution, but if you add in an explanation for how you came up with that, I'll be sure to explain why it doesn't work.
