### Instruction: Benjamin is trying to break a combination lock. This particular type of lock has 5 digits from 0 to 9, and Benjamin just happens to know that there can be no repeated digits in the code for this type of lock. How many valid codes are there?### Response: The first thing I notice is that this is a permutation problem, since the order of the digits matters.
I also notice that the digits cannot be repeated, which means that each digit I choose eliminates a possibility for the next digit.
So, to count the number of valid codes, I need to multiply the number of choices I have for each digit, starting from the leftmost one.
For the leftmost digit, I have 10 choices, since any digit from 0 to 9 can be used.
For the second digit, I have 9 choices, since I cannot use the digit I already chose for the first digit.
Similarly, for the third digit, I have 8 choices, for the fourth digit, I have 7 choices, and for the fifth digit, I have 6 choices.
Therefore, the total number of valid codes is 10 times 9 times 8 times 7 times 6, which is equal to 30240.