Number of car plates having $3$ letters and a $4$-digit number. Car plates have are to be designed by placing $3$ (not necessarily distinct) English letters followed by a $4$-digit number, where the digits of this number can not be simultaneously $0$, that is, the string "$0000$" is excluded. Also, $0$ can be used in the leftmost place of the number.
How many such car plates are there?
For many years, I was thinking that the answer is:
$26 \times 26 \times 26 \times 9999 = 175,742,424$

BUT
Suddenly I have a doubt about that. I think, in that way of calculation, we did not consider the permutation of the letters and numbers.

To clarify, take this as an example:
In how many ways can we arrange the letters if the word "MATHEMATICS"?
The answer is $\frac{11!}{2!2!2!}=6,652,800$ because we have $2$ M's, $2$ A's, and $2$ T's.
So, the word "MATHEMATICS" and the word "MATHEMATICS" are apparently same, except that the $2$ M's exchanged their positions. So we need not over count. Hence we divide by $2!$, and so on.

Going back to the original problem, I have doubt that we are over counting.
For instance, "BHB $1035$" is counted twice in the $175,742,424$ ways because of the "B"

I need your clarification regarding this. Thanks.
 A: Your original answer is correct.
There are $26$ choices for each letter, giving $26^3$ ways to fill the three positions allocated for letters.  Similarly, if we initially ignore the constraint that the final four positions cannot be filled with $0000$, we would have $10$ choices for each digit, giving $10^4$ ways to fill the four positions allocated for digits.  Since we cannot use $0000$, one string of digits is eliminated, leaving us with $$26^3(10^4 - 1)$$ possible license plates.
Since you are concerned about repetitions, let's count the license plates a different way.
First, we will count the number of ways of filling the first three positions with letters.
Three distinct letters:  There are $26$ ways to fill the first letter, $25$ ways to fill the second letter, and $24$ ways to fill the third letter.  Hence, there are $$26 \cdot 25 \cdot 24$$ ways to fill the first three positions with distinct letters.
Exactly two distinct letters:  Since we have three positions to fill, one letter must appear twice and another letter must appear once.  There are $26$ ways to choose the letter which appears twice, $25$ ways to choose the letter which appears once, and three ways to choose the position of the letter which appears only once.  Hence, there are $$26 \cdot 25 \cdot 3$$ ways to fill the first three positions with exactly two distinct letters.
One letter fills all three positions:  There are $26$ ways to choose the letter which fills all three positions.
Total:  Since these three cases are mutually exclusive and exhaustive, the number of ways we can fill the first three positions with letters is
$$26 \cdot 25 \cdot 24 + 26 \cdot 25 \cdot 3 + 26 = 26^3$$
Next, we will count the number of ways of filling the last four positions with digits.
Four distinct digits:  There are $10$ ways to fill the first digit, $9$ ways to fill the second digit, $8$ ways to fill the third digit, and $7$ ways to fill the fourth digit, so there are $$10 \cdot 9 \cdot 8 \cdot 7$$ ways to fill these positions with four distinct digits.
Exactly three distinct digits:  Since we have four positions to fill, one digit must appear twice and two other digits must each appear once.  There are $10$ ways to select the digit which appears twice, $\binom{4}{2}$ ways to choose two of the four positions for that digit, $\binom{9}{2}$ ways to select two of the remaining nine digits to each appear once, and $2!$ ways to arrange those digits in the remaining two positions.  Hence, there are $$\binom{10}{1}\binom{4}{2}\binom{9}{2}2!$$
ways to fill these positions with exactly three distinct digits.
Exactly two distinct digits:  Either one digit appears three times and another digit appears once or two digits each appear twice.
One digit appears three times and another digit appears once:  There are $10$ ways to choose the digit which appears three times, $9$ ways to choose the digit which appears once, and $4$ ways to choose the position of the digit which appears exactly once.  Hence, there are $$10 \cdot 9 \cdot 4$$ such choices.
Two digits each appear twice:  There are $\binom{10}{2}$ ways to select the two digits which will each appear twice and $\binom{4}{2}$ ways to select two of the four positions for the smaller of those digits.  Hence, there are $$\binom{10}{2}\binom{4}{2}$$ such choices.
One digit appears all four positions:  Ignoring the constraint that we cannot use $0000$ gives us $10$ choices for the digit which fills all four positions.  Since we have that constraint, we actually have just $9$ choices.
Total:  Since these cases are mutually exclusive and exhaustive, the number of admissible ways to fill the last four positions with digits is
$$10 \cdot 9 \cdot 8 \cdot 7 + \binom{10}{1}\binom{4}{2}\binom{9}{2}2! + 10 \cdot 9 \cdot 4 + \binom{10}{2}\binom{4}{2} + 9 = 10^4 - 1$$
Since we can choose letters and digits independently, the number of license plates we can form is indeed $$26^3(10^4 - 1)$$ which demonstrates that your original approach was correct.
A: I believe that your initial thought is right. Let's take BHB $1035$ as an example:
If we are forming a random license plate, we have $1/26$ probability for the first letter to be "B", $1/26$ probability for the second to be "H" and $1/26$ probability for the third to be "B". Also we have $1/9999$ probability for the number at the end to be $1035$. Thus, the probability of a random license plate number to be "BHB $1035$" is $\dfrac{1}{26\cdot 26\cdot 26\cdot9999}$.
Another way to look at this might be as follows:
We find the number of possible 4-digit combinations as $10\cdot 10 \cdot 10 \cdot 10$, and excluding "$0000$", $10^4-1 = 9999$. You may solve for the number of possible 3-letter combinations in the same fashion, as the difference is just that there are $26$ different choices for one place instead of $10$.
A: Your original concept is correct, and your 2nd concept, which focuses on permutations is wrong.  This is because permutations are appropriate when you are sampling (letters and/or numbers) without replacement.
In the present problem, other than the constraint against $...0000$, all letters and numbers are sampled with replacement.  This means that a license plate such as "AAA1111" for example, is acceptable.  So, you do have $26$ independent choices for each letter, and $10$ independent choices for each number (minus the not "...0000" constraint).
