License plate combinations 4 letters 3 numbers How many license plates can be created if they have $4$ letters and $3$ digits in any order?
My thinking is $26^4 \times 10^3 \times {7 \choose 4} \times {7 \choose 3}$:
$26$ possibilities for each of 4 letters, $10$ possibilities for each of 3 digits, $4$ places out of $7$ to place the letters, and $3$ places out of $7$ to place the numbers.
Is this correct?
Reiterating what was said earlier in a comment:
Your answer is almost correct however you made the mistake of picking the positions twice rather than only once.  By choosing the four positions to be occupied by letters and then again choosing which three positions were to be occupied by digits and having chosen said positions out of the original seven, you will have run the risk of picking the same position multiple times designating it to be simultaneously used by letter and digit.
Instead, the digits' positions should have been selected out of the remaining unused positions giving a total of $\binom{3}{3}=1$ way in which this step could be accomplished, this being merely $1$ way it can be left out of the final product.  This leaves the final answer as being:
$$26^4\cdot 10^3\cdot \binom{7}{4}$$

Compare this to a slightly larger problem where you have $26$ letters to choose from, $10$ digits, and $15$ special symbols (e.g. !@#\$%^ and so on...) and we want to make a ten-character password consisting of $6$ letters, $3$ digits, and $1$ special symbol with repetition allowed.
We first pick the positions used by letters, then from those remaining positions pick the positions used by digits, letting the final remaining position be occupied by the symbol, and then picking which letters, digits, and symbol they happened to be for a total count of:
$$26^6\cdot 10^3\cdot 15\cdot \binom{10}{6}\cdot \binom{4}{3}$$
There are 28 letters and 10 digts from 0 to 9. You want to chose 7 of them when the order in which you select is important. So there are $P_{38}^7$ possibilites. ($P$ stands for permutation)