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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 $3$ numbers from $1$ to $20$ . If no numbers in the code are allowed to repeat, how many different codes could be made? $\newline$ Answer:

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Compound events: find the number of outcomes

Full solution

Q. 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?

\newline

Answer:

Permutation Calculation: To solve this problem, we need to calculate the number of ways to choose $3$ different numbers from $20$ without repetition. This is a permutation problem because the order of the numbers matters for the lock code. $\newline$ Calculation: The first number can be any of the $20$ numbers. For the second number, we have $19$ options (since we can't repeat the first number). For the third number, we have $18$ options (since we can't repeat the first or second numbers). $\newline$ So the total number of different codes is $20 \times 19 \times 18$ .
Total Number Calculation: Now we perform the multiplication to find the total number of different codes. $\newline$ Calculation: $20 \times 19 \times 18 = 6840$

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