If a lock requires 3 different digits from 0 to 9, how many different combinations can be formed?

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Q: If a lock requires 3 different digits from 0 to 9, how many different combinations can be formed?

Solution: The number of ways to choose 3 different digits from 10 is 10P3 = 720.

Steps: 6

Step 1: Understand that we need to choose 3 different digits from the numbers 0 to 9. This means we have 10 possible digits to choose from.
Step 2: Recognize that the order in which we choose the digits matters because they will form a combination for the lock.
Step 3: Use the formula for permutations to find the number of ways to arrange 3 digits out of 10. The formula is nPr = n! / (n - r)!, where n is the total number of items to choose from, and r is the number of items to choose.
Step 4: In our case, n = 10 (the digits 0 to 9) and r = 3 (the digits we want to choose). So we calculate 10P3.
Step 5: Calculate 10P3: 10! / (10 - 3)! = 10! / 7! = (10 × 9 × 8) = 720.
Step 6: Conclude that there are 720 different combinations of 3 different digits that can be formed.