Question

question
a lock has a 3 - number code made up of 12 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?

Explanation:

Step1: Identify the problem type

This is a permutation problem since the order of the numbers in the code matters (it's a unique code with order), and we are choosing 3 different numbers from 12 without repetition. The formula for permutations is \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 12 \) (total numbers) and \( r = 3 \) (numbers to choose).

Step2: Apply the permutation formula

Substitute \( n = 12 \) and \( r = 3 \) into the formula:
\( P(12, 3)=\frac{12!}{(12 - 3)!}=\frac{12!}{9!} \)
Since \( n!=n\times(n - 1)\times\cdots\times1 \), we can simplify \( \frac{12!}{9!}=\frac{12\times11\times10\times9!}{9!}=12\times11\times10 \)

Step3: Calculate the result

\( 12\times11\times10 = 1320 \)

Answer:

1320