Number of possible outcomes in a license plate

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If a license plate consists of 3 letters followed by 3 digits and having at least one digit or letter repeated .. How many outcomes are there?

26 * 26 * 10* 10 * 10 .. Is that right?

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3 Answers

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That is incorrect.

First, let's find the number of license tags, total, that can be made with $3$ letters and $3$ digits. This is: $$26\cdot26\cdot26\cdot10\cdot10\cdot10 = 26^310^3$$

Second, look at how many tags can be made without repeating any character (letter or number): $$26\cdot25\cdot24\cdot10\cdot9\cdot8$$

Now, if we must repeat at least one character, then number of tags that satisfy are: $$\text{Total} - \text{ThoseThatDon'tRepeat} = 26^310^3-26\cdot25\cdot24\cdot10\cdot9\cdot8$$

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The total number of possibilities is $A = 26^3 \times 10^3$. The number of possibilities with no repeats is $B = 26 \times 25 \times 24 \times 10 \times 9 \times 8$. So the number of possibilities with at least one repeat is just $A-B$.

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Let us count the three letter strings with at least one letter repeated. There are $26$ where all three letters are the same. To have two letters the same and one different, you have $26$ choices for the double letter, $25$ choices for the single letter, and $3$ choices for the location of the single letter, so there are $26 \cdot 25 \cdot 3 + 26=1976$ total strings with at least one letter repeated. I'll let you count the three digit strings with at least one repeat-call the number $N$. The logic is the same.

Then we would be tempted to say the final answer is $1976\cdot 10^3 + 26^3N$, because if we have a repeated letter we can pair that set with any string of digits. Unfortunately, we have double counted the cases where you have a repeated letter and a repeated digit, so the final answer is $1976\cdot 10^3 + 26^3N- 1976N$ by the inclusion-exclusion principle

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