If a lock requires a 3-digit code using the digits 0-9, how many different codes

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Question: If a lock requires a 3-digit code using the digits 0-9, how many different codes can be formed if digits cannot be repeated?

Options:

Correct Answer: 720

Solution:

The first digit has 10 options, the second has 9, and the third has 8. Total = 10 * 9 * 8 = 720.

If a lock requires a 3-digit code using the digits 0-9, how many different codes can be formed if digits cannot be repeated?

If a lock requires a 3-digit code using the digits 0-9, how many different codes can be formed if digits cannot be repeated?

Step 1: Understand that we need to create a 3-digit code using the digits from 0 to 9.
Step 2: Realize that there are 10 different digits available (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
Step 3: For the first digit of the code, you can choose any of the 10 digits. So, there are 10 options.
Step 4: For the second digit, you cannot use the digit you already chose for the first digit. This leaves you with 9 options.
Step 5: For the third digit, you cannot use the first or second digit. This leaves you with 8 options.
Step 6: To find the total number of different codes, multiply the number of options for each digit together: 10 (first digit) * 9 (second digit) * 8 (third digit).
Step 7: Calculate the total: 10 * 9 * 8 = 720.

Permutations – The question tests the understanding of permutations where the order matters and digits cannot be repeated.
Counting Principles – It assesses the ability to apply the fundamental counting principle to determine the total number of combinations.