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A lock has a 3-number code made up of 30 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 $30$ 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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Compound events: find the number of outcomes

Full solution

Q. A lock has a

3

-number code made up of

30

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 combinations possible for a $3$ -number code from $30$ unique numbers without repetition. This is a permutation problem because the order of the numbers matters for the code. $\newline$ To calculate the number of permutations of $n$ items taken $r$ at a time, we use the formula: $\newline$ $P(n, r) = \frac{n!}{(n - r)!}$ $\newline$ In this case, $n = 30$ (total numbers) and $r = 3$ (numbers in the code).
Calculate Factorial of n: First, we calculate the factorial of $n$ , which is $30!$ ( $30$ 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 with the $(n - r)!$ in the denominator. So we calculate $30 \times 29 \times 28$ .
Calculate Product of First $r$ Terms: The calculation for $30 \times 29 \times 28$ gives us: $\newline$ $30 \times 29 \times 28 = 24,360$ $\newline$ This is the number of unique $3$ -number codes that can be made from $30$ numbers without repetition.

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