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.
Best Answer
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.