Question

a lock has a code of 4 numbers from 1 to 10. if no numbers in the code are allowed to repeat, how many different codes could be made? answer attempt 1 out of 2

Explanation:

Step1: Identify the problem type

This is a permutation problem since the order of numbers in the code matters (different order means different code) and no repetition is allowed. The formula for permutations of \( n \) objects taken \( r \) at a time is \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 10 \) (numbers from 1 to 10) and \( r = 4 \) (code length).

Step2: Calculate the permutation

First, calculate \( n! = 10! = 10\times9\times8\times7\times6\times5\times4\times3\times2\times1 \), and \( (n - r)!=(10 - 4)!=6! = 6\times5\times4\times3\times2\times1 \). Then, \( P(10, 4)=\frac{10!}{6!}=\frac{10\times9\times8\times7\times6!}{6!}=10\times9\times8\times7 \).

Step3: Compute the product

\( 10\times9 = 90 \), \( 90\times8 = 720 \), \( 720\times7 = 5040 \).

Answer:

5040