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A lock has a code of 3 numbers from 1 to 20 . If no numbers in the code are allowed to repeat, how many different codes could be made? Answer:

A lock has a code of 33 numbers from 11 to 2020 . If no numbers in the code are allowed to repeat, how many different codes could be made?\newlineAnswer:

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Full solution

Q. A lock has a code of 33 numbers from 11 to 2020 . If no numbers in the code are allowed to repeat, how many different codes could be made?\newlineAnswer:
  1. Permutation Calculation: To solve this problem, we need to calculate the number of ways to choose 33 different numbers from 2020 without repetition. This is a permutation problem because the order of the numbers matters for the lock code.\newlineCalculation: The first number can be any of the 2020 numbers. For the second number, we have 1919 options (since we can't repeat the first number). For the third number, we have 1818 options (since we can't repeat the first or second numbers).\newlineSo the total number of different codes is 20×19×1820 \times 19 \times 18.
  2. Total Number Calculation: Now we perform the multiplication to find the total number of different codes.\newlineCalculation: 20×19×18=684020 \times 19 \times 18 = 6840

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