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A lock has a 3-number code made up of 23 numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code?
Answer:

A lock has a $3$ -number code made up of $23$ numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code? $\newline$ Answer:

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Experimental probability

Full solution

Q. A lock has a

3

-number code made up of

23

numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code?

\newline

Answer:

Calculate Permutations Formula: We need to calculate the number of ways to choose $3$ different numbers from $23$ without repetition for a lock code. This is a permutation problem because the order of the numbers matters. $\newline$ To calculate the number of permutations of $n$ items taken $r$ at a time without repetition, we use the formula: $\newline$ $P(n, r) = \frac{n!}{(n - r)!}$ $\newline$ In our case, $n = 23$ (total numbers available) and $r = 3$ (numbers to choose for the code).
Calculate Factorial of $n$ : First, we calculate the factorial of $n$ , which is $23!$ ( $23$ factorial). However, we don't need to calculate the entire factorial. We only need to calculate the product of the first $r$ terms of $n$ , because the rest will cancel out when we divide by $(n - r)!$ . So, we calculate $23 \times 22 \times 21$ .
Calculate Factorial of $(n - r)$ : Now, we calculate the factorial of $(n - r)$ , which is $(23 - 3)!$ or $20!$ . Again, we don't need to calculate the entire factorial because it will cancel out with the terms in $23!$ .
Divide to Find Permutations: We divide the product of the first $r$ terms of $n$ by $(n - r)!$ to get the number of permutations. $\newline$ $P(23, 3) = \frac{23 \times 22 \times 21}{20!}$ $\newline$ Since $20!$ is in both the numerator and the denominator, they cancel each other out. $\newline$ $P(23, 3) = 23 \times 22 \times 21$
Perform Multiplication: We perform the multiplication to find the number of permutations. $P(23, 3) = 23 \times 22 \times 21 = 10626$
Final Result: We have found the number of different ways to choose three different numbers from $23$ for a unique $3$ -number lock code without repeating any numbers.

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